Quantum Mechanics Problems David J. Jeffery

Quantum Mechanics Problems

David J. Jeffery

Department of Physics & Health Physics

Idaho State University

Pocatello, Idaho

♠

♠ ♠ ♠ ♠

♠ ♠

♠

♠ ♠

Portpentagram Publishing (self-published)

2001 January 1

Introduction

Quantum Mechanics Problems (QMP) is a source book for instructors of introductory quantummechanics. The book is available in electronic form to instructors by request to the author. It is freecourseware and can be freely used and distributed, but not used for commercial purposes. The aimof QMP is to provide digestable problems for quizzes, assignments, and tests for modern students.There is a bit of spoon-finding—nourishing spoon-feeding I hope.

The problems are grouped by topics in chapters: see Contents below. The chapter orderingfollows roughly the traditional chapter/topic ordering in quantum mechanics books. For each chapterthere are two classes of problems: in order of appearance in a chapter they are: (1) multiple-choiceproblems and (2) full-answer problems. Almost all the problems have complete suggested answers.The answers may be the greatest benefit of QMP. The questions and answers can be posted on theweb in pdf format.

The problems have been suggested by many sources, but have all been written by me. Giventhat the ideas for problems are the common coin of the realm, I prefer to call my versions of theproblems redactions. Instructors, however, might well wish to find solutions to particular problemsfrom well known texts. Therefore, I give the suggesting source (when there is one or when I recallwhat it was) by a reference code on the extra keyword line: e.g., (Gr-10:1.1) stands for Griffiths,p. 10, problem 1.1. Caveat: my redaction and the suggesting source problem will not in generalcorrespond perfectly or even closely in some cases. The references for the source texts and otherreferences follow the contents. A general citation is usually, e.g., Ar-400 for Arfken, p. 400.

At the end of the book are three appendices. The first is set of review problems anent matricesand determinants. The second is an equation sheet suitable to give to students as a test aid and areview sheet. The third is a set of answer tables for multiple choice questions.

Quantum Mechanics Problems is a book in progress. There are gaps in the coverage and theordering of the problems by chapters is not yet final. User instructors can, of course, add and modifyas they list.

Everything is written in plain TEX in my own idiosyncratic style. The questions are all havecodes and keywords for easy selection electronically or by hand. A fortran program for selectingthe problems and outputting them in quiz, assignment, and test formats is also available. Note thequiz, etc. creation procedure is a bit clonky, but it works. User instructors could easily constructtheir own programs for problem selection.

I would like to thank the Department of Physics & Health Physics of Idaho State Universityfor its support for this work. Thanks also to the students who helped flight-test the problems.

Contents

1 Classical Physics in Trouble2 QM Postulates, Schrodinger Equation, and the Wave Function3 Infinite Square Wells and Other Simple Wells4 The Simple Harmonic Oscillator (SHO)5 Free Particles and Momemtum Representation6 Foray into Advanced Classical Mechanics7 Linear Algebra8 Operators, Hermitian Operators, Braket Formalism9 Time Evolution and the Heisenberg Representation

10 Measurement11 The Central Force Problem and Orbital Angular Momentum

i

12 The Hydrogenic Atom13 General Theory of Angular Momentum14 Spin15 Time-Independent Approximation Methods16 Time-Dependent Perturbation17 The Hydrogenic Atom and Spin18 Symmetrization Principle19 Atoms20 Molecules21 Solids22 Interaction of Radiation and Matter23 Second Quantization24 Klein-Gordon Equation25 Ephemeral Problems

Appendices

1 Mathematical Problems2 Quantum Mechanics Equation Sheet3 Multiple-Choice Problem Answer Tables

References

Abramowitz, M., & Stegun, I. A. 1972, Handbook of Mathematical Functions with Formulas,

Graphs,s and Mathematical Tables (Washington, D.C.: U.S. Government Printing Office)(AS)

Adler, R., Bazin, M., & Schiffer, M. 1975, Introduction to General Relativity (New York:McGraw-Hill Book Company) (ABS)

Arfken, G. 1970, Mathematical Methods for Physicists (New York: Academic Press) (Ar)Baym, G. 1969, Lectures on

Quantum Mechanics (Reading, Massachusetts: Benjamin/Cummings Publishing Co., Inc.)(Ba)

Behte, H., & Jackiw, R. W. 1986, Intermediate Quantum Mechanics (Menlo Park, California:Benjamin/Cummings Publishing Company) (BJ)

Bernstein, J., Fishbane, P. M., & Gasiorowicz, S. 2000, Modern Physics (Upper Saddle River,New Jersey: Prentice Hall) (BFG)

Bevington, P. R. 1969, Data Reduction and Error Analysis for the Physical Sciences (New-York:McGraw-Hill Book Company) (Bev)

Cohen-Tannoudji, C., Diu, B., & Laloe, F. 1977, Quantum Mechanics (New York: John Wiley& Sons) (Co)

Chandrasekhar, S. 1960, Radiative Transfer (New York: Dover Publications, Inc.) (Ch)Dahl, J. P. 2001, Introduction to the Quantum World of Atoms and Molecules (Singapore: World

Scientific) (Da)Davisson, C. M. 1965, in Alpha-, Beta-, and Gamma-Ray Spectroscopy, ed. K. Siegbahn

(Amsterdam: North-Holland), 37 (Da)Enge, H. A. 1966, Introduction to Nuclear Physics (Reading, Massachusetts: Addison-Wesley

Publishing Company, Inc.) (En)Gasiorowicz, S. 1974, Quantum Physics (New York: John Wiley & Sons) (Ga)Greiner, W. 1994, Quantum Mechanics: An Introduction (Berlin: Springer-Verlag) (Gre)Griffiths, D. J. 1995, Introduction to Quantum Mechanics (Upper Saddle River, New Jersey:

Prentice Hall) (Gr)Griffiths, D. J. 2005, Introduction to Quantum Mechanics (Upper Saddle River, New Jersey:

Prentice Hall) (Gr2005)

ii

Goldstein, H., Poole, C. P., Jr., & Safko, J. L. 2002, Classical Mechanics, 3rd Edition (SanFrancisco: Addison-Wesley Publishing Company) (GPS)

Harrison, W. A. 2000, Applied Quantum Mechanics (Singapore: World Scientific) (Ha)Hodgman, C. D. 1959, CRC Standard Mathematical Tables, 12th Edition (Cleveland, Ohio:

Chemical Rubber Publishing Company) (Hod)Jackson, J. D. 1975, Classical Electrodynamics (New York: John Wiley & Sons)Jeffery, D. J. 2001, Mathematical Tables (Port Colborne, Canada: Portpentragam Publishing)

(MAT)Keenan, C. W., Wood, J. H., & Kleinfelter, D. C., 1976, General College Chemistry (New York:

Harper & Row, Publishers) (Ke)Leighton, R. B. 1959, Principles of Modern Physics (New York: McGraw-Hill Book Company,

Inc.) (Le)Mihalas, D. 1978, Stellar Atmospheres, 2nd Edition (San Francisco: W. H. Freeman and

Company) (Mi)Morrison, M. A. 1990, Understanding Quantum Mechanics (Upper Saddle River, New Jersey:

Prentice Hall) (Mo)Morrison, M. A., Estle, T. L., & Lane, N. F. 1991, Understanding More Quantum Mechanics

(Upper Saddle River, New Jersey: Prentice Hall) (MEL)Pathria, R. K. 1980, Statistical Mechanics (Oxford: Pergamon Press) (Pa)Pointon, A. J. 1967, Introduction to Statistical Physics (London: Longman Group Ltd.) (Po)Tipler, P. A. 1978, Modern Physics (New York: Worth Publishers, Inc.) (Ti)

iii

Chapt. 1 Classical Physics in Trouble

Multiple-Choice Problems

001 qmult 00100 1 1 3 easy memory: quantum mechanics

1. The physical theory that deals mainly with microscopic phenomena is:

a) quartz mechanics.

b) quarks mechanics.

c) quantum mechanics.

d) quantum jump mechanics.

e) quasi-mechanics.

001 qmult 00200 1 1 1 easy memory: photon energy

2. The photon, the quantum of electromagnetic radiation, has ENERGY:

a) hf = h−ω.

b) h/λ.

c) h−k.d) h2f .

e) hf2.

001 qmult 00300 1 1 4 easy memory: photoelectric effect

3. A key piece of evidence for the wave-particle duality of light is:

a) the photograph effect.

b) the Maxwell’s electrodynamics as summarized in the four Maxwell’s equations.

c) the frequency of red light.

d) the photoelectric effect.

e) the photomagnetic effect.

001 qmult 00400 1 1 1 easy memory: Compton effect

4. Einstein predicted and Compton proved that photons:

a) have linear momentum.

b) do not have linear momentum.

c) sometimes have linear momentum.

d) both have and do not have linear momentum at the same time.

e) neither have nor have not linear momentum.

001 qmult 00500 1 4 3 easy deducto-memory: Bohr atom

5. “Let’s play Jeopardy! For $100, the answer is: This model of an atom is of historical andpedagogical interest, but it is of little use in modern practical calculations and from the modernstandpoint is probably misleading rather than insight-giving.”

What is , Alex?

1

2 Chapt. 1 Classical Physics in Trouble

a) Schrodinger’s model of the hydrogen atomb) the Thomas-Fermi model of a many electron atomc) Bohr’s model of the hydrogen atom d) the liquid drop model of the atome) the model hydrogen atom of Leucippos and Democritos

001 qmult 00550 1 1 4 easy memory: hydrogenic energy formula6. The formula

En = −1

2mec

2α2Z2

n2

gives the main energy levels of:

a) positronium.b) magnesium deboride.c) the hydrogen molecule.d) the hydrogenic atom.e) the infinite square well.

001 qmult 00600 1 1 5 easy memory: Greek atomists7. The atomic theory was first proposed by the ancient Greeks Leucippos (5th century BCE) and

Democritos (5th to 4th century BCE: he reputedly lived to be 100). The term atomos meansuncut: e.g., the grass is atomos. The atomists started from a philosophical position that therehad to be something to give stability to nature: obviously the macroscopic world was full ofchange: therefore what was imperishable or uncutable—atoms—must be below perception. Themodern quantum theory does indeed bear out some of their thinking. Microscopic particles canbe created and destroyed, of course, but the members of a class are much more identical thanmacroscopic objects can ever be: fundamental particles like electrons and quarks are thoughtto be absolutely identical. Thus the forms particles can take are apparently eternal: a hydrogenatom today is the same in theory as one at any time in universal history.

The atomists tried to work out an atomic understanding of existence in general. Forinstance they constructed a cosmology using atoms that bears some resemblance to moderneternal inflationary cosmology in which there are infinitely many universes that are born outof primordial space-time foam and perhaps return to that—foam to foam. Unfortunately, theatomists got off on the wrong foot on the shape of the Earth: they were still flat Earthers whenthe round Earth theory was being established. Quite obviously to us, the atomists were badlynon-experimental. Much of their thinking can be called rational myth. To a degree they werelucky in happening to be attracted to an essentially right idea.

The atomists were eventually stigmatized as atheists: they did not deny that gods exist, butdidn’t leave anything for the gods to do. This may have been their downfall. The more orthodoxand popular philosophies of Plato, Aristotle, and the Stoics rejected atomism probably, amongother things, for its seeming atheism. Christianity followed suit in this regard. The writings ofthe atomists only exist in fragments—and Democritos seems to have been as famous as Platoin his day. The Epicurean philosophers adopted atomism, but also suffered the stigmatizationas atheists—and also hedonists who are, of course, the worst. But the atom idea lingered onthrough the centuries: Leucippos and Democritos, Epicurus, Lucretius (his surviving poem De

Rerum Natura [On Nature] expounds atomism), Gassendi (17th century), Newton, Dalton: thechain is unbroken: it is not true that modern atomism has no historical or essential connectionto ancient atomism.

A good account of ancient atomism can be found in David Furley’s The Greek Cosmologists.

Now, without recurring to the top of this preamble, atomism was invented in:

a) the early 19th century. b) the 17th century by Gassendi.c) the 10th century CE. d) the 5th century CE. e) the 5th century BCE.

Chapt. 1 Classical Physics in Trouble 3

001 qmult 00800 1 1 1 easy memory: causality, relativity8. Einstein ruled out faster than light signaling because:

a) it would cause irresolvable causality paradoxes.b) it would not cause irresolvable causality paradoxes.c) it led to irresolvable paradoxes in quantum mechanics.d) it would destroy the universe.e) it had been experimentally verified.

001 qmult 00900 1 1 3 easy memory: EPR paradox9. The Einstein-Podolsky-Rosen (EPR) paradox was proposed to show that ordinary quantum

mechanics implied superluminal signaling and therefore was:

a) more or less correct.b) absolutely correct.c) defective.d) wrong in all its predictions.e) never wrong in its predictions.

001 qmult 01000 1 4 3 easy deducto-memory: Bell’s theorem10. “Let’s play Jeopardy! For $100, the answer is: This theorem (if it is indeed inescapably correct)

and the subsequent experiments on the effect the theorem dealt with show that quantummechanical signaling exceeds the speed of light.”

a) What is Dark’s theorem, Alex?b) What is Midnight’s theorem, Alex?c) What is Bell’s theorem, Alex?d) What is Book’s theorem, Alex?e) What is Candle’s theorem, Alex?

Full-Answer Problems

001 qfull 00500 3 5 0 tough thinking: Rutherford’s nucleusExtra keywords: (HRW-977:62P)

1. Rutherford discovered the nucleus in 1911 by bombarding metal foils with alpha particles nowknown to be helium nuclei (atomic mass 4.0026). An alpha particle has positive charge 2e. Heexpected the alpha particles to pass right through the foils with only small deviations. Most did,but some scattered off a very large angles. Using a classical particle picture of the alpha particlesand the entities they were scattering off of he came to the conclusion that atoms contained mostof their mass and positive charge inside a region with a size scale of ∼ 10−15 m = 1 fm: this 10−5

times smaller than the atomic size. (Note fm stands officially for femtometer, but physicistscall this unit a fermi.) Rutherford concluded that there must be a dense little core to an atom:the nucleus.

a) Why did the alpha particles scatter off the nucleus, but not off the electrons? HINTS:Think dense core and diffuse cloud. What is the force causing the scattering?

b) If the alpha particles have kinetic energy 7.5 Mev, what is their de Broglie wavelength?

c) The closest approach of the alpha particles to the nucleus was of order 30 fm. Would thewave nature of the alpha particles have had any effect? Note the wave-particle duality wasnot even suspected for massive particles in 1911.


001 qfull 01000 3 5 0 tough thinking: black-body radiation, Wien lawExtra keywords: (Le-62) gives a sketch of the derivations

2. Black-body radiation posed a considerable challenge to classical physics which it was partiallyable to meet. Let’s see how far we can get from a classical, or at least semi-classical,thermodynamic equilibrium analysis.

a) Let Uλ be the radiation energy density per wavelength of a thermodynamic equilibriumradiation field trapped in some kind of cavity. The adjective thermodynamic equilibriumimplies that the field is homogenous and isotropic. I think Hohlraum was the traditionalname for such a cavity. Let’s call the field a photon gas and be done with it—anachronismbe darned. Since the radiation field is isotropic, the specific intensity is then given by

B(λ, T ) =cUλ

4π, (Pr.1)

where c is of course the speed of light. Specific intensity is radiation flux per wavelengthper solid angle. From special relativity (although there may be some legitimately classicalway of getting it), the momentum flux associated with a specific intensity is just B(λ, T )/c.Recall the rest plus kinetic energy of a particle is given by

E =√

p2c2 +m20c

4 , (Pr.2)

where p is momentum and m0 is rest mass. From an integral find the expression for theradiation pressure on a specularly reflecting surface:

p =1

3U , (Pr.3)

where p is now pressure and U is the wavelength-integrated radiation density. Argue thatthe same pressure applies even if the surface is only partially reflecting or pure blackbodyprovided the the radiation field and the surface are in thermodynamic equilibrium. HINT:Remember to account for angle of incidence and reflection.

b) Now we can utilize a few classical thermodynamic results to show that

U = aT 4 , (Pr.4)

where a is a radiation constant related to the Stefan-Boltzmann constant σ = 5.67051 ×105 ergs/(cm2 K4) and T is Kelvin temperature, of course. The relation between a and σfollows from the find the flux leaking out a small hole in the Hohlraum:

F = 2π

∫ 1

0

cU

4πµ dµ =

ca

4T 4 , (Pr.5)

where µ is the cosine of the angle to the normal of the surface where the hole is. One seesthat σ = ca/4. Classically a cannot be calculated theoretically; in quantum mechanicalstatistical mechanics a can be derived. The proportionality U ∝ T 4 can, however, bederived classically. Recall the 1st law of thermodynamics:

dE = T dS − p dV , (Pr.6)

where E is internal energy, S is entropy, and V is volume. Note that

(

∂E

∂S

)

V

= T and

(

∂E

∂V

)

S

= −p , (Pr.7)


where the subscripts indicate the variables held constant. It follows from calculus (assumingwell-behaved functions) that

(

∂p

∂S

)

V

= −(

∂T

∂V

)

S

, (Pr.8)

The last relation is one of Maxwell’s four thermodynamic relations—Newton did things inthrees; Maxwell in fours. Note that E = UV for a radiation field. Now go to it: showU ∝ T 4.

c) As a by-product of the part (b) answer, you should have found that

T ∝ V −1/3 (Pr.9)

for a quasistatic adiabatic process with the photon gas. (Find it now if somehow you missedit in the part (b) answer.) Assume you have a perfectly reflecting Hohlraum that you expandhomologously by a scaling factor f(t), where t is time. Thus at any time t any length ℓ betweenphysical points on the walls in the system is given by

ℓ = f(t)ℓ0 , (Pr.10)

where ℓ0 was the physical length at t0 when f(t0) = 1. Find out how T , U , U dV , and E scalewith f(t). What happens to the lost internal energy? HINT: This is easy.

d) Consider the process described in the part (c) and show that

λ = λ0f(t) (Pr.11)

for each specific intensity beam. Note you can use the non-relativistic Doppler effect since thevelocity shift between scatterings off the walls is vanishingly small in the quasistatic limit.

e) For the same system as in part (c) show that

B(λ, T ) dλ dV = f(t)−1B(λ0, T0) dλ0 dV0 . (Pr.12)

Then show that equation (Pr.12) leads naturally (if not absolutely necessarily so far as I cansee) to the prescription for black-body specific intensity

B(λ, T ) = λ−5g(x) =

(

T

x

)5

g(x) , (Pr.13)

wherex ≡ λT (Pr.14)

and g(x) is a universal function that cannot be determined from classical theory.Equation (Pr.13) is sometimes called Wien’s displacement law. However the name Wien’sdisplacement law is more usually (I think) reserved for the immediate result that for fixed Tthe the maximum of the black-body specific intensity (i.e., the maximum of x−5g(x)) occurs ata wavelength given by

λ =xmax

T, (Pr.15)

where xmax is the global universal location of maximum for the universal function g(x). It wasempirically known that black-body radiation had only one maximum with wavelength, and sothis corresponds to xmax. I think classically xmax has to be determined empirically.

Wien’s radiation law was I believe a fit to the observations of Wien’s displacement law.This law is

B(λ, T ) = k1

(

T

x

)5

exp

(

−k2

x

)

, (Pr.16)


where k1 and k2 had to be determined from the fit. Wien’s law works well for short wavelengths(x <∼xmax), but gives a poorish fit to the long wavelengths (x >∼xmax) (Pa-190, but note the xthere is the inverse of the x here aside from a constant). The Rayleigh-Jeans law derived froma rather different classical starting picture gave a good fit to long wavelengths (x >> xmax),but failed badly at shorter wavelengths (Pa-190, but note the x there is the inverse of the xhere aside from a constant). In fact the Rayleigh-Jeans law goes to inifinity as x goes to zeroand the total energy in a Rayleigh-Jeans radiation field is infinite (Le-64): this is sometimescalled the ultraviolet catastrophe (BFG-106). The correct black-body specific intensity law wasderived from a primitive quantum theory by Max Planck in 1900 (BFG-106). Planck obtainedan empirically excellent fit to the black-body specific intensity and then was able to derive itfrom his quantum hypothesis. The Rayleigh-Jeans and Planck laws are the subject for anotherquestion.

001 qfull 01100 2 5 0 moderate thinking: Bohr atom3. In 1913, Niels Bohr presented his model of the hydrogen atom which was quickly generalized

to the hydrogenic atom (i.e., the one-electron atom of any nuclear charge Z). This modelcorrectly gives the main hydrogenic atom energy levels and consists of a mixture of quantummechanical and classical ideas. It is historically important for showing that quantization issomehow important in atomic structure and pedagogically it is of interest since it shows howsimple theorizing can be done. But the model is, in fact, incorrect and from the modernperspective probably even misleading about the quantum mechanical nature of the atom. It ispartially an accident of nature that it exists to be found. Only partially an accident since itdoes contain correct ingredients.

And it is no accident that Bohr found it. Bohr knew what he wanted: a model that wouldsuccessfully predict the hydrogen atom spectrum which is a line spectrum showing emissionat fixed frequencies. He knew from Einstein’s photoelectric effect theory that electromagneticradiation energy was quantized in amounts hν where h = 6.62606896(33) × 10−27 erg s wasPlanck’s constant (which was introduced along with the quantization notion to explain black-body radiation in 1900) and ν was frequency of the quantum of radiation. He recognized thatPlanck’s constant had units of angular momentum. He knew from Rutherford’s nuclear modelof the atom that the positive charge of an atom was concentrated in region that was muchsmaller than the atom size and that almost all the mass of the atom was in the nucleus. Heknew that there were negative electrons in atoms and they were much less massive than thenucleus. He knew the structure of atoms was stable somehow. By a judicious mixture ofclassical electromagnetism, classical dynamics, and quantum ideas he found his model. A moresophisticated mixture of these concepts would lead to modern quantum mechanics.

Let’s see if we can follow the steps of the ideal Bohr—not the Bohr of history.NOTE: This a semi-classical question: Bohr, ideal or otherwise, knew nothing of theSchrodinger equation in 1913. Also note that this question uses Gaussian CGS units not MKSunits. The most relevant distinction is that electric charge

eCGS =eMKS√4πǫ0

which implies the fine structure constant in CGS is

α =e2

h−c.

Astronomy is all Gaussian CGS by the way.

a) Bohr thought to build the electron system about the nucleus based on the electrostaticinverse square law with the electron system supported against collapse onto the nucleus byangular momentum. The nucleus was known to be much tinnier than the electron system


which gives the atom its volume. The nucleus could thus be a considered an immobilepoint center of force at the origin of the relative nucleus-electron coordinate system frame.This frame is non-inertial, but classically can be given an inertial-frame treatment if theelectron is given a reduced mass given by

m =memnucleus

me +mnucleus≈ me

(

1 − me

mnucleus

)

,

where me the electron mass and mnucleus is the nucleus mass. The approximation is validfor me/mnucleus << 1 which is true of hydrogen and most hydrogenic systems, but not, forexample, for positronium (a bound electron and positron).The electron—there is only one in a hydrogenic atom—was taken to be in orbit about thenucleus. Circular orbits seemed the simplest way to proceed. The electrostatic force law(in Gaussian cgs units) in scalar form for a circular orbit is

~F = −Ze2

r2r ,

where Ze is the nuclear charge, e is the electron charge, and r is the radial distance to theelectron, and r is a unit vector in the radial direction.

What is the potential energy of the electron with the zero of potential energy for theelectron at infinity as usual? HINT: If the result isn’t obvious, you can get it using thework-potential energy formula:

V = −∫

~F · d~r + constant .

b) Using the centripetal force law (which is really F = ma for uniform circular motion)

~F = −mv2

rr ,

find an expression for the classical kinetic energy T of the electron in terms of Z, e, and ralone.

c) What is the total energy of the electron in the orbit?

d) Classically an accelerating charge radiates. This seemed well established experimentally inBohr’s time. But an orbiting electron is accelerating, and so should lose energy continuouslyuntil it collapses into the nucleus: this catastrophe obviously doesn’t happen. Electronsdo not collapse into the nucleus. Also they radiate only at fixed frequencies which meansfixed quantum energies by Einstein’s photoelectric effect theory. So Bohr postulated thatthe electron could only be in certain orbits which he called stationary states and that theelectron in a stationary state did not radiate. Only on transitions between stationary states(sometimes called quantum jumps or leaps) was there an emission of radiation in a quantumof radiation or (to use an anachronism) a photon. To get the fixed energies of emissiononly certain energies were allowed for the stationary states. But the emitted photons didn’tcome out with equally spaced energies: ergo the orbits couldn’t be equally spaced in energy.From the fact that Planck’s constant h has units of angular momentum, Bohr hypothesizedthe orbits were quantized in equally spaced amounts of angular momentum. But h was notthe spacing that worked. Probably after a bit of fooling around, Bohr found that h/(2π)or, as we now write it, h− was the spacing that gave the right answer. The allowed angularmomenta were given by

L = nh− ,


where n is any positive non-zero integer. The n is now called the principal quantum number,but its meanings in the Bohr model and in modern quantum mechanics are somewhatdifferent. The principal quantum number n determines the main spacing of the hydrogenicenergy levels.

Rewrite kinetic energy T in terms of nh− and solve for an expression for r in termsn, h−, Ze2, and m only. HINT: Recall the classical expression for angular momentum ofparticle in a circular orbit is L = mrv.

e) Using the formula for r from the part (d) answer write an expression for the energy of astationary state in terms of m, c, α, Z, and n only. The c is the speed of light and the αis the fine structure constant: recall that in Gaussian cgs units

α =e2

h−c.

This formula for orbit energy turns out to be correct for the spacing of the main energylevels. But these energyxhell doesn’t, in fact, have angular momentum nh−: it consists ofhas orbitals (as we now call them) with angular momenta in the range [0, n− 1] in units ofh− (e.g., Gr-139).

001 qfull 01300 2 3 0 moderate math: Compton scatteringExtra keywords: (Ha-323:1.1)

4. In 1916, Einstein proposed that photons carry momentum according to the following formula:

p =h

λ,

where h is Planck’s constant and λ is the photon wavelength (HRW-959). In 1924, Louisde Broglie applied the formula in inverse form to give a wavelength for massive particles: i.e.,

λ =h

p

which is called the de Broglie wavelength formula. In 1923, Arthur Compton carried outexperiments with X-rays scattering off electrons which showed that Einstein’s formula correctlyaccounted for the wavelength shift on scattering. The Compton shift formula is

∆λ = λC(1 − cos θ) ,

where λC = h/(mec) = 0.02426 A is the Compton wavelength (with me being the electron mass)and θ is the scattering angle (i.e., the angle between incident and scattering directions). Thisformula can be derived from Einstein’s formula using a relativistic particle collisional picture.

a) Assuming an electron starts at rest and is hit head-on by a photon “particle and thecollision is elastic,” what conservation law expressions can be used to relate incomingphoton momentum p1, outgoing photon momentum p2, outgoing electron momentum pe,photon scattering angle θ, and electron scattering angle φ? Can one solve for the fouroutgoing quantities given the initial conditions? HINT: Recall that the relativistic kineticenergy of a particle is given by

T =√

(pc)2 + (m0c2)2 −m0c2 = (γ − 1)m0c

2 ,

where p is momentum and m0 is the rest mass.

b) Solve for p2 in terms of p1 and θ only.

c) Now using Einstein wavelength formula, find Compton’s formula.


d) Sketch the behavior of ∆λ as a function of θ. What is the shift formula in the non-relativisticlimit: i.e., when λ→ ∞.

001 qfull 00150 3 5 0 tough thinking: Einstein, RunyonExtra keywords: Bosher

5. “God does not play dice”—Einstein. Discuss.

Chapt. 2 QM Postulates, Schrodinger Equation, and the Wave Function


002 qmult 00080 1 1 2 easy memory: wave-particle duality1. The nebulous (and sometimes disparaged) concept that all microscopic physical entities have

both wave and particle properties is called the wave-particle:

a) singularity. b) duality. c) triality. d) infinality. e) nullility.

002 qmult 00090 1 4 5 easy deducto-memory: Sch eqn2. “Let’s play Jeopardy! For $100, the answer is: The equation that governs (or equations

that govern) the time evolution of quantum mechanical systems in the non-relativisticapproximation.”

What is/are , Alex?

a) ~Fnet = m~a b) Maxwell’s equationsc) Einstein’s field equations of general relativity d) Dirac’s equatione) Schrodinger’s equation

002 qmult 00100 1 1 1 easy memory: Sch eqn compact form3. The full Schrodinger’s equation in compact form is:

a) HΨ = ih−∂Ψ

∂t. b) HΨ = h−∂Ψ

∂t. c) HΨ = i

∂Ψ

∂t. d) HΨ = ih−∂Ψ

∂x.

e) H−1Ψ = ih−∂Ψ

∂t.

002 qmult 00110 1 1 3 easy memory: Hamiltonian operator4. The energy operator in quantum mechanics,

H = − h−2

2m

∂2

∂x2+ V (x)

(here given for 1 particle in one dimension) is called the:

a) Lagrangian b) Laplacian c) Hamiltonian d) Georgian e) Torontonian

002 qmult 00200 1 4 3 easy deducto-memory: Born postulateExtra keywords: mathematical physics

5. “Let’s play Jeopardy! For $100, the answer is: The postulate that the wave function Ψ(~r ) isquantum mechanics is a probability amplitude and |Ψ(~r )|2 is a probability density for localizinga particle at ~r on a ‘measurement’.”

What is , Alex?

a) Schrodinger’s idea b) Einstein’s notion c) Born’s postulated) Dirac’s hypothesis e) Death’s conclusion

002 qmult 00210 1 1 1 easy memory: QM probability density6. In the probabilistic interpretation of wave function Ψ, the quantity |Ψ|2 is:

10

Chapt. 2 QM Postulates, Schrodinger Equation, and the Wave Function 11

a) a probability density. b) a probability amplitude. c) 1. d) 0.e) a negative probability.

002 qmult 00220 1 1 5 easy memory: probability of finding particle in dx7. The probability of finding a particle in differential region dx is:

a) Ψ(x, t) dx. b) Ψ(x, t)∗ dx. c) [Ψ(x, t)∗/Ψ(x, t)] dx. d) Ψ(x, t)2 dx.e) Ψ(x, t)∗Ψ(x, t) dx = |Ψ(x, t)|2 dx.

002 qmult 00300 1 4 5 easy deducto-memory: observable definedExtra keywords: See Co-137, Gr-104

8. “Let’s play Jeopardy! For $100, the answer is: It is an Hermitian operator that governs (orrepresents in some people’s jargon) a dynamical variable in quantum mechanics.”

What is an , Alex?

a) intangible b) intaglio c) obtainable d) oblivion e) observable

002 qmult 00310 1 1 3 easy memory: expectation value defined9. In quantum mechanics, a dynamical variable is governed by a Hermitian operator called an

observable that has an expectation value that is:

a) the most likely value of the quantity given by the probability density: i.e., the mode of theprobability density.

b) the median value of the quantity given by the probability density.c) the mean value of the quantity given by the probability density.d) any value you happen to measure.e) the time average of the quantity.

002 qmult 00320 1 1 3 easy memory: expectation value notation10. The expectation value of operator Q for some wave function is often written:

a) Q. b) 〉Q〈. c) 〈Q〉. d) 〈f(Q)〉. e) f(Q).

002 qmult 00400 1 1 1 easy memory: physical requirmentsExtra keywords: Gr-11

11. These quantum mechanical entities must be (with some exceptions):

i) Single-valued (and their derivatives too).ii) finite (and their derivatives too).iii) continuous (and their derivatives too).iv) normalizable or square-integrable.

They are:a) wave functions. b) observables. c) expectation values. d) wavelengths.e) wavenumbers.

002 qmult 00410 1 1 4 easy memory: normalization requirement12. A physical requirement on wave functions is that they should be:

a) reliable. b) friable. c) certifiable. d) normalizable. e) retriable.

002 qmult 00500 1 1 2 easy memory: the momentum operator defined13. The momentum operator in one-dimension is:

a) h− ∂

∂x. b)

h−i

∂

∂x. c)

i

h−∂

∂x. d)

i

h−∂

∂t. e) h− ∂

∂t.

002 qmult 00510 1 1 4 easy memory: constant of the motion14. If an observable has no explicit time dependence and it commutes with the Hamiltonian, then

it is a quantum mechanical:

12 Chapt. 2 QM Postulates, Schrodinger Equation, and the Wave Function

a) fudge factor. b) dynamical variable. c) universal constant.d) constant of the motion. e) constant of the stagnation.

002 qmult 00520 1 4 5 easy deducto-memory: Ehrenfest’s theorem15. Ehrenfest’s theorem partially shows the connection between quantum mechanics and:

a) photonics. b) electronics. c) special relativity. d) general relativity.e) classical mechanics.

002 qmult 00600 1 4 5 easy deducto-memory: uncertainty principle 116. “Let’s play Jeopardy! For $100, the answer is: It describes a fundamental limitation on the

accuracy with which we can know position and momentum simultaneously.”

What is , Alex?

a) Tarkovsky’s doubtful thesis b) Rublev’s ambiguous postulatec) Kelvin’s nebulous zeroth law d) Schrodinger’s wild hypothesise) Heisenberg’s uncertainty principle

002 qmult 00610 1 4 5 easy deducto-memory: uncertainty principle 217. “Let’s play Jeopardy! For $100, the answer is: ∆x∆p ≥ h−/2 or σxσp ≥ h−/2.

What is , Alex?

a) an equality b) a standard deviationc) the Heisenberg CERTAINTY principle d) the Cosmological principlee) the Heisenberg UNCERTAINTY principle

002 qmult 00700 1 1 4 easy memory: Schr. eqn. separation of variables18. The time-independent Schrodinger equation is obtained from the full Schrodinger equation by:

a) colloquialism. b) solution for eigenfunctions.c) separation of the x and y variables. d) separation of the space and time variables.e) expansion.

002 qmult 00720 1 1 1 easy memory: stationary state19. A system in a stationary state will:

a) not evolve in time. b) evolve in time. c) both evolve and not evolve in time.d) occasionally evolve in time. e) violate the Heisenberg uncertainty principle.

002 qmult 00800 1 4 2 easy deducto-memory: orthogonality property20. For a Hermitian operator eigenproblem, one can always find (subject to some qualitifications

perhaps—but which are just mathemtical hemming and hawwing) a complete set (or basis) ofeigenfunctions that are:

a) independent of the x-coordinate. b) orthonormal. c) collinear.d) pathological. e) righteous.

002 qmult 00810 1 4 2 easy deducto-memory: basis expansionExtra keywords: mathematical physics

21. “Let’s play Jeopardy! For $100, the answer is: If it shares the same same range as a basis setof functions and is at least piecewise continuous, then it can be expanded in the basis with avanishing limit of the mean square error between it and the expansion.”

What is a/an , Alex?

a) equation b) function c) triangle d) deduction e) tax deduction

002 qmult 00820 1 4 5 easy deducto-memory: general Born postulateExtra keywords: mathematical physics


22. “Let’s play Jeopardy! For $100, the answer is: The postulate that expansion coefficients ofa wave function in the eigenstates of an observable are the probability amplitudes for wavefunction collapse to eigenstates of that observable.”

What is , Alex?

a) the special Born postulate b) the very special Born postulatec) normalizability d) the mass-energy equivalence e) the general Born postulate

002 qmult 00830 1 1 4 easy memory: basis expansion physics23. The expansion of a wave function in an observable’s basis (or complete set of eigenstates) is

a) just a mathematical decomposition. b) useless in quantum mechanics.c) irrelevant in quantum mechanics. d) not just a mathematical decomposition sincethe expansion coefficients are probability amplitudes. e) just.

020 qmult 00840 1 4 5 easy deducto-memory: wave function collapseExtra keywords: mathematical physics

24. “Let’s play Jeopardy! For $100, the answer is: It is a process in quantum mechanics that somedecline to mention, some believe to be unspeakable, some believe does not exist (though theyhave got some explaining to do about how one ever measures anything), some believe shouldnot exist, and that some call the fundamental perturbation (but just once per textbook).”

What is , Alex?

a) the Holy b) the Unholy c) the Unnameabled) the 4th secret of the inner circle e) wave function collapse

002 qmult 00900 1 4 1 easy deducto-memory: macro object in stationary state25. “Let’s play Jeopardy! For $100, the answer is: A state that no macroscopic system can be

in except arguably for states of Bose-Einstein condensates, superconductors, superfluids andmaybe others sort of.”


a) stationary state b) accelerating state c) state of the Union d) state of beinge) state of mind

002 qmult 01000 1 1 5 easy memory: stationary state is radical26. A stationary state is:

a) just a special kind of classical state. b) more or less a kind of classical state.c) voluntarily a classical state.d) was originally not a classical state, but grew into one.e) radically unlike a classical state.

002 qmult 01100 1 1 4 easy memory: macro system in a stationary state27. Except arguably for certain special cases (superconductors, superfluids, and Bose-Einstein

condensates), no macroscopic system can be in a:

a) mixed state. b) vastly mixed state. c) classical state. d) stationary state.e) state of the union.

002 qmult 01200 1 1 2 easy memory: transitions28. Transitions between atomic or molecular stationary states (sometimes, but actually rarely, called

quantum jumps) are:

a) only collisional.b) both collisional and radiative.c) only radiative.


d) neither collisional nor radiative.e) only collisional to higher energy stationary states and only radiative to lower energy

stationary states.

002 qmult 01300 1 4 3 easy deducto-memory: lasers, stimulated emission29. “Let’s play Jeopardy! For $100, the answer is: It is the basis for lasers and masers.”

What is , Alex?

a) spontaneous radiative emission b) desultory radiative emissionc) stimulated radiative emission d) the laser force e) the laser potential

002 qmult 01400 1 4 4 easy deducto-memory: operators and Sch. eqn.30. “Let’s play Jeopardy! For $100, the answer is: An equation that must hold in order for the non-

relativistic Hamiltonian operator and the operator ih−∂/∂t to both yield an energy expectationvalue for a wave function Ψ(x, t).”

What is , Alex?

a) the continuity equation b) the Laplace equation c) Newton’s 2nd lawd) Schrodinger’s equation e) Hamiton’s equation

002 qmult 02000 2 1 4 moderate memory: does gravity quantizeExtra keywords: reference: Nesvizhevsky et al. 2002, Nature, 413, 297

31. Can the gravitational potential cause quantization of energy states?

a) No. b) It is completely uncertain. c) Theoretically yes, but experimentally no.d) Experimental evidence to date (post-2001) suggests it can.e) In principle there is no way of telling.


002 qfull 00090 1 5 0 easy thinking: what is a wave function?1. What is a wave function? (Representative general symbol Ψ(~r, t)).

002 qfull 00100 1 3 0 easy math: probability and age distributionExtra keywords: (Gr-10:1.1)

2. Given the following age distribution, compute its the normalization (i.e., the factor thatnormalizes the distribution), mean, variance, and standard deviation. Also give the mode(i.e., the age with highest frequency) and median. HINT: Doing the calculation with a smallcomputer code would be the efficient way to answer the problem.

Table: Age Distribution

Age Frequency(years)

14 215 116 622 224 225 5


002 qfull 00200 2 3 0 moderate math: probability needle 1Extra keywords: (Gr-10:1.3) probability and continuous variables

3. An indicator needle on a semi-circular scale (e.g., like a needle on car speedometer) bouncesaround and comes to rest with equal probability at any angle θ in the interval [0, π].

a) Give the probability density ρ(θ) and sketch a plot of it.

b) Compute the 1st and 2nd moments of the distribution (i.e., 〈θ〉 and⟨

θ2⟩

) and the varianceand standard deviation.

c) Compute 〈sin θ〉, 〈cos θ〉, 〈sin2 θ〉, and 〈cos2 θ〉.

002 qfull 00210 3 5 0 tough thinking: 2-variable probability densityExtra keywords: (Gr-11:1.5) dropping a needle on lines needle 2

4. Nun fur eine kleine teufelische problem. Say you drop at random with equal likelihood of landingin any orientation and location a needle of length ℓ onto a sheet of paper with parallel lines adistance ℓ apart. What is the probability of the needle crossing (or at least touching) a line?Let’s be nice this time and break it down.

a) Mentally mark one end of needle red. Then note that really we only need to consider oneband on the paper between two parallel lines and the case where the red end lies betweenthem as a given. Why is this so?

b) So now we consider that the red end lands in one band at a point x between −ℓ/2 and ℓ/2.Note we put the origin at the center since almost always one ought to exploit symmetry.What is the probability density for the red end to land anywhere in the band? What isthe probability density for the needle for the orientation of the needle in θ measured fromthe x-axis? Why do you only need to consider θ ∈ [0, π]?

c) Now we don’t care about the orientation itself really: we just care about it’s projectionon the x-axis. Call that projection x′. What is the probability density for x′? What isthe range of x′ allowed? HINT: The probability of landing in dθ and a corresponding dx′

must be equal.

d) The joint probability density for x and x′ is

ρ(x)ρ(x′) .

You now have to integrate up all the probability for x′ + x ≥ ℓ/2 and for x′ + x ≤ −ℓ/2and sum those two probabilities. The sum is the solution probability of course.

002 qfull 00220 1 3 0 easy math: Gaussian probability densityExtra keywords: (Gr-11:1.6)

5. Consider the Gaussian probability density

ρ(x) = Ae−λ(x−a)2 ,

where A, a, and λ are constants.

a) Determine the normalization constant A.

b) The nth moment of a probability density is defined by

〈xn〉 =

∫ ∞

−∞

xnρ(x) dx .

Determine the 0th, 1st, and 2nd moments of the Gaussian probability density.


c) For the Gaussian probability density determine the mean, mode, mediam, variance σ2, andstandard deviation (or dispersion) σ.

d) Sketch the Gaussian probability density.

002 qfull 00300 2 3 0 moderate math: analyzing a triangular hat wave functionExtra keywords: (Gr-13:1.7)

6. At some time a triangular hat wave function is given by

Ψ(x, t) =

Ax

a, x ∈ [0, a];

A

(

b− x

b− a

)

, x ∈ [a, b];

0 otherwise,

where A, a, and b are constants.

a) Sketch Ψ and locate most probable location for a particle (i.e., the mode of the |Ψ|2probability distribution).

b) Determine the normalization constant A in terms of a and b. Recall the difference betweenwave function and probability distribution here and in the later parts of this question.

c) What are the probabilities of being found left and right of a, respectively?

d) What is 〈x〉?

002 qfull 00310 2 5 0 moderate thinking: probability conservationExtra keywords: (Gr-13:1.9) probability current

7. The expression for the probability that a particle is in the region [−∞, x] (i.e., the cumulativeprobability distribution function) is

P (x, t) =

∫ x

−∞

|Ψ(x′, t)|2 dx′ .

a) Find an explicit, non-integral expression for ∂P (x, t)/∂t given that the wave function isnormalizable at time t. HINT: Make use of the physics: i.e., the Schrodinger equationitself. This is a common trick in quantum mechanics and, mutatis mutandis, throughoutphysics.

b) If the wave function is normalizable at time t, show that P (∞, t) is a constant with respectto time: i.e., total probability is conserved.

c) The probability current is defined

J(x, t) = −∂P (x, t)

∂t.

Argue that this is a sensible definition.

d) GivenΨ(x, t) = ψ(x)e−iωt ,

what can one say about the probability density |Ψ|2, the cumulative probability functionP (x, t), and the probability current J(x, t)?

002 qfull 00320 3 5 0 tough thinking: general time evolution equation


8. It follows from the general Born postulate that the expectation value of an observable Q is givenby

〈Q〉 =

∫ ∞

−∞

Ψ∗QΨ dx .

It’s weird to call an operator an observable, but that is the convention (Co-137).

a) Write down the explicit expression for

d〈Q〉dt

.

Recall Q in general can depend on time too.

b) Now use the Schrodinger equation

HΨ = ih−∂Ψ

∂t

to eliminate partial time derivatives where possible in the expression for d〈Q〉/dt.Remember how complex values behave when complex conjugated. You should use theangle bracket form for expectation values to simplify the expression where possible.

c) The commutator of two operators A and B is defined by

[A,B] = AB −BA ,

where it is always understood that the commutator and operators are acting on an implicitgeneral function to the right. If you have trouble initially remembering the understoodcondition, you can write

[A,B]f = (AB −BA)f ,

where f is an explicit general function. Operators don’t in general commute: i.e.,[A,B] = AB −BA 6= 0 in general. Prove

∑

i

Ai,∑

j

Bj

=∑

i,j

[Ai, Bj ] .

d) Now show that d〈Q〉/dt can be written in terms of 〈i[H,Q]〉. The resulting importantexpression oddly enough doesn’t seem to have a common name. I just call it the generaltime evolution formula. HINTS: First, V and Ψ∗ do commute. Second, the other part ofthe Hamiltonian operator

T = − h−2

2m

∂2

∂x2

can be put in the right place using integration by parts and the normalization conditionon the wave function. Note T turns out to be the kinetic energy operator.

e) If d〈Q〉/dt = 0, then Q is a quantum mechanical constant of the motion. It’s weird to callan observable (which is a operator) a constant of the motion, but that is the convention(Co-247). Show that the operator Q = 1 (i.e., the unit operator) is a constant of themotion. What is 〈1〉?

f) Find the expression for d〈x〉/dt in terms of what we are led to postulate as the momentumoperator

p =h−i

∂

∂x.


The position operator x should be eliminated from the expression. HINTS: Note V andx commute, but T and x do not. Leibniz’s formula (Ar-558) might be of use in evaluatingthe commutator [T, x]. The formula is

dn(fg)

dxn=

n∑

k=0

(

n

k

)

dkf

dxk

dn−kg

dxn−k.

002 qfull 00330 3 5 0 tough thinking: Ehrenfest’s theoremExtra keywords: (Gr-17:1.12) Ehrenfest formulae

9. In one dimension, Ehrenfest’s theorem in quantum mechanics is usually taken to consist of twoformulae:

d〈x〉dt

=1

m〈p〉

andd〈p〉dt

= −⟨

∂V

∂x

⟩

,

where the angle brackets indicate expectation values as usual.

a) From the general time evolution formula prove the 1st Ehrenfest formula. HINTS: Recall thegeneral time evolution formula in non-relativistic quantum mechanics is

d〈Q〉dt

=

⟨

∂Q

∂t

⟩

+1

h−〈i[H,Q]〉 ,

where Q is any observable and H is the Hamiltonian:

H = T + V (x) .

Also recall that quantum mechanical momentum operator and kinetic energy operator are givenby

p =h−i

∂

∂xand T = − h−2

2m

∂2

∂x2,

respectively. Leibniz’s formula (Ar-558) might be of use in evaluating some of the commutators:

dn(fg)

dxn=

n∑

k=0

(

n

k

)

dkf

dxk

dn−kg

dxn−k.

b) From the general time evolution formula prove the 2nd Ehrenfest formula.

c) In the macroscopic limit, the expectation values become the classical dynamical variables bythe correspondence principle (which is an auxiliary principle of quantum mechanics enunciatedby Bohr in 1920 (Wikipedia: Correspondance principle)): i.e., 〈x〉 becomes x, etc. (Note we areallowing a common ambiguity in notation: x and p are both coordinates and, in the classicalformalism, the dynamical variables describing the particle. Everybody does this: who are we dodisagree.) Find the macroscopic limits of the Ehrenfest formulae and identify the macroscopiclimits in the terminology of classical physics.

d) If you ARE writing a TEST, omit this part.If one combines the two Ehrenfest formulae, one gets

md2〈x〉dt2

= −⟨

∂V

∂x

⟩


which looks very like Newton’s 2nd law in its F = ma form for a force given by a potential. Usingthe correspondence priniciple, it does become the 2nd law in the macroscopic limit. However,an interesting question arises—well maybe not all that interesing—does the 〈x〉 (which we couldcall the center of the wave packet) actually obey the 2nd law-like expression

md2〈x〉dt2

= −∂V (〈x〉)∂〈x〉 ?

To disprove a general statement, all you need to do is find one counterexample. Consider apotential of the form V (x) = Axλ, and show that in general the 〈x〉 doesn’t obey 2nd law-likeexpression given above. Then show that it does in three special cases of λ.

002 qfull 00400 2 3 0 moderate math: orthonormality leads to mean energyExtra keywords: (Gr-30:2.10)

10. You are given a complete set of orthonormal stationary states (i.e., energy eigenfunctions)ψn and a general wave equation Ψ(x, t) that is for the same system as ψn: i.e., Ψ(x, t) isdetemined by the same Hamiltonian as ψn. The set of eigen-energies of ψn are En. Thesystem is bounded in space by x = −∞ and x = ∞.

a) Give the formal expansion expression of Ψ(x, 0) (i.e., Ψ(x, t) at time zero) in terms of ψn.Also give the formal expression for the coefficients of expansion cn.

b) Now give the formal expansion for Ψ(x, t) remembering that ωn = En/h−. Justify that thisis the solution of the Schrodinger equation for the initial conditions Ψ(x, 0).

c) Find the general expression, simplified as far as possible, for expectation value 〈Hℓ〉 interms of the expansion coefficients, where ℓ is any positive (or zero) integer. Are thesevalues time dependent?

d) Give the special cases for ℓ = 0, 1, and 2, and the expression for the standard deviation forenergy σE . HINTS: This should be a very short answer: 3 or 4 lines.

002 qfull 00500 3 5 0 tough thinking: real eigen-energiesExtra keywords: (Gr-24:2.1) and all real complete sets

11. There are a few simple theorems one can prove about stationary states and their eigen-energies.

a) Prove that eigen-energies must be real. HINT: Prove 〈H〉 is real for any state Ψ usingintegration by parts. Note one has to use the full time dependent wave function for a generalstate since the time dependence doesn’t cancel out of the expectation value integral.

b) The complete set of time-independent stationary states you get from a direct solution ofthe Schrodinger equation may not be all pure real pure. But one can always construct fromthis complete set another complete set that is all pure real and it is supposedly convenientto do so sometimes—or at least it can be done as a mathematician would say. Show howit can be done. HINTS: First note that complete sets are almost always assumed tobe minimum complete sets: i.e., each member of the set is independent of all the othermembers, and thus cannot be constructed from any linear combination of the others. Inour discussions we always assume minimum complete sets.

Consider a non-trivially complex solution ψij of the eigenproblem

Hψij = Eiψij ,

where the first subscript denotes energy level and the second the particular solution of thatenergy level. (“Non-trivially” just means that ψij isn’t just a real function times a complexconstant. What do you do with a trivially complex ψij by the way?) Take the complexconjugate of the eigenproblem to find an independent 2nd solution ψ2nd to it with the same


energy. The 2nd solution may or may not be part of your original subset with energy Ei.If it is, then that is good. But if it isn’t one of the original subset with energy Ei, youshould replace one of those with ψ2nd. Since the original set was complete

ψ2nd =∑

ℓ

cℓψiℓ ,

where the summation only needs to run over the eigenfunctions with the same energy Ei.This equation can be rearranged for any ψim (except for ψij itself):

ψim =∑

ℓ

cℓ 6=mψiℓ + cψ2nd ,

where the coefficients cℓ all had to be changed and c is the coefficient needed for ψ2nd.Since ψim can be constructed using ψ2nd, it can be replaced by ψ2nd. If the number ofstates with energy Ei is infinite, the replacement process becomes hairy, but let’s not worryabout that.

Now construct two pure real solutions from ψij and ψ2nd from which ψij and ψ2nd

can be re-constructed. These two new states then replace ψij and ψ2nd in the subset withenergy Ei. One can go on like that replacing two for two as long as you need to. Rememberthe original set will in general be infinite, and one couldn’t have had them all explicitlyanyway.

002 qfull 00600 3 5 0 tough thinking: parity operator12. The parity operator P (not to be confused with the momentum operator p) has the well defined,

but seemingly arbitrary, property that

Pf(x) = f(−x)

for a 1-dimensional case which is all that we will consider in this problem.

a) Prove the parity operator is Hermitian. HINTS: Recall that the definition of the Hermitianconjugate of operator Q is

〈φ|Q|ψ〉 = 〈ψ|Q†|φ〉∗ ,where |φ〉 and |ψ〉 are arbitrary kets. Note Q is Hermitian if Q† = Q. Since the parityoperator (as defined here) only has meaning in the position representation that is wherethe proof must be done: thus one must prove

∫ ∞

−∞

φ(x)∗Pψ(x) dx =

[∫ ∞

−∞

ψ(x)∗Pφ(x) dx

]∗

.

A transformation of the integration variable might help: remember x in the integrals isjust a dummy variable that can be represented by any symbol.

b) The eigenproblem for the parity operator is

Pf(x) = pvalf(x) ,

where pval are the eigenvalues. Solve for the complete set of eigenvalues and identify thoseclasses of functions which are eigenfunctions of P . HINTS: Note it’s f(x) on the righthand side not f(−x) since this is an eigenproblem, but Pf(x) = f(−x) too. Recall thatthe eigenvalues of a Hermitian operator are pure real. Nothing forbids using the parityoperator twice. The parity operator commutes with constants of course:

P [cf(x)] = cf(−x) = cPf(x) .


c) The set of all eigenfunctions of P is complete. Thus P qualifies as an “observable” in QMjargon whether it can be observed or not: i.e., it is a Hermitian operator with a completeset of eigenstates. Show that the set of eigenstates is complete: i.e., that any functionf(x) can be written in an expansion of P eigenfunctions. HINTS: From any f(x) onecan construct another function f(−x) and from f(x) and f(−x) one can construct twoeigenfunctions of P , and from those two eigenfunctions of P one can reconstruct . . .

d) If f ′(x) is the derivative of f(x), then Pf ′(x) = f ′(−x): i.e., the derivative of f(x) evaluatedat −x. But what is

∂

∂x[Pf(x)] ?

Do P and ∂/∂x commute? Do P and ∂2/∂x2 commute? HINT: You’ve heard of the chainrule.

e) If the potential is even (i.e., V (x) = V (−x)) do P and the Hamiltonian H commute?HINTS: Recall PV (x)f(x) must be interpreted in QM (unless otherwise clarified) as Pacting on the function V (x)f(x) not on V (x) alone.

f) Given that P and H commute and ψ(x) is a solution of the time-independent Schrodingerequation, show that ψ(−x) a solution too with the same eigen-energy as ψ(x): i.e., ψ(x)and ψ(−x) are degenerate eigenstates.

g) Given that P and H commute, show how one can construct from a given complete setof energy eigenstates a complete set of energy eigenstates that are also eigenstates of theparity operator. Assume that the original complete set contains both ψ(x) and ψ(−x):this is not a requirement for finding a common complete set, but it is a simplification here.HINT: Recall the part (c) answer.

002 qfull 01000 2 5 0 moderate thinking: energy and normalizationExtra keywords: (Gr-24:2.2) zero-point energy

13. Classically E ≥ Vmin for a particle in a conservative system.

a) Show that this classical result must be so. HINT: This shouldn’t be a from-first-principlesproof: it should be about one line.

b) The quantum mechanical analog is almost the same: E = 〈H〉 > Vmin for any normalizablestate of the system considered. Note the equality E = 〈H〉 = Vmin never holds quantummechanically. (There is an over-idealized exception, which we consider in part (e).) Provethe inequality. HINTS: The key point is to show that 〈T 〉 > 0 for all physically allowedstates. Use integration by parts.

c) Now show that result E > Vmin implies E > Vmin, where E is any eigen-energy of thesystem considered. Note the equality E = Vmin never holds quantum mechanically (exceptfor the over-idealized system considered in part (e)). In a sense, there is no rest state forquantum mechanical particle. This lowest energy is called the zero-point energy.

d) The E > Vmin result for an eigen-energy in turn implies a 3rd result: any ideal measurementalways yields an energy greater than Vmin Prove this by reference to a quantum mechanicalpostulate.

e) This part is NOT to be done on EXAMS: it’s just too much (for the grader). Thereis actually an exception to E > Vmin result for an eigen-energy where E = Vmin occurs.The exception is for quantum mechanical systems with periodic boundary conditions and aconstant potential. In ordinary 3-dimensional Euclidean space, the periodic boundaryconditions can only occur for rings (1-dimensional systems) and sphere surfaces (2-dimensional systems) I believe. Since any real system must have a finite size in all 3spatial dimensions, one cannot have real systems with only periodic boundary conditions.


Thus, the exception to the E > Vmin result is for unrealistic over-idealized systems. Let usconsider the idealized ring system as an example case. The Hamiltonian for a 1-dimensionalring with a constant potential is

H = − h−2

2mr2∂2

∂φ2+ V ,

where r is the ring radius, φ is the azimuthal angle, and V is the constant potential. Findthe eigen-functions and eigen-energies for the Schrodinger equation for the ring systemwith periodic boundary conditions imposed. Why must one impose periodic boundaryconditions on the solutions? What solution has eigen-energy E = Vmin?

002 qfull 00110 2 5 0 moderate thinking: beyond the classical turning points14. The constant energy of a classical particle in a conservative system is given by

E = T + V .

Since classically T ≥ 0 always, a bound particle is confined by surface defined by T = 0 orE = V (~r ). The points constituting this surface are called the turning points: a name whichmakes most sense in one dimension. Except for static cases where the turning point is triviallythe rest point (and maybe some other weird cases), the particle comes to rest only for aninstant at a turning point since the forces are unbalanced there. So it’s a place where a particle“ponders for an instant before deciding where to go next”. The region with V > E is classicallyforbidden. Now for most quantum mechanical potential wells, the wave function extends beyondthe classical turning point surface into the classical forbidden zone and, in fact, usually goes tozero only at infinity. If the potential becomes infinite somewhere (which is an idealization ofcourse), the wave function goes to zero: this happens for the infinite square well for instance.

Let’s write the 1-dimensional time-independent Schrodinger equation in the form

∂2ψ

∂x2=

2m

h−2 (V − E)ψ .

a) Now solve for ψ for the region with V > E with simplifying the assumption that V isconstant in this region.

b) Can the solutions be normalized?

c) Can the solutions constitute an entire wave function? Can they be part of a wave function?In which regions?

d) Although we assumed constant V , what crudely is the behavior of the wave function likelyto be like the regions with V > E.

e) For typical potentials considered at our level, qualitatively what is the likelihood of findingthe particle in the classically forbidden region? Why?

002 qfull 01100 3 5 0 tough thinking: 1-d non-degeneracy15. If there are no internal degrees of freedom (e.g., spin) and they are NORMALIZABLE, then

one-particle, 1-dimensional energy eigenstates are non-degenerate. We (that is to say you) willprove this.

Actually, we know already that any 2nd order ordinary linear differential equation has onlytwo linearly independent solutions (Ar-402) which means, in fact, that from the start we knowthere is a degeneracy of 2 at most. Degeneracy count is the number of independent solutions.If there is more than one independent solution, then infinitely many linear combinations ofsolutions have the same energy. But in an expansion of wave function, only a set linearindependent solutions is needed and thus the number of such solutions is the physically relevant


degeneracy. Of course, our proof means that one of the linearly independent solutions is notnormalizable.

a) Assume you have two degenerate 1-dimensional energy eigenstates for Hamiltonian H : ψ1

and ψ2. Prove that ψ1ψ′2 − ψ2ψ

′1 equals a constant where the primes indicate derivative

with respect to x the spatial variable. HINT: Write down the eigenproblem for both ψ1

and ψ2 and do some multiplying and subtraction and integration.

b) Prove that the constant in part (a) result must be zero. HINT: To be physically allowableeigenstates, the eigenstates must be normalizable.

c) Integrate the result of the part (b) answer and show that the two assumed solutions arenot physically distinct. Show for all x that

ψ2(x) = Cψ1(x) ,

where C is a constant. This completes the proof of non-degeneracy since eigenstates thatdiffer by a multiplicative constant are not physically (i.e., expansion) distinct. HINT:You have to show that there is no other way than having ψ2(x) = Cψ1(x) to satisfythe condition found in the part (b) answer. Remember the eigenproblem is a linear,homogeneous differential equation.

002 qfull 01200 2 3 0 mod math: 3-d exponential wave function, probabilityExtra keywords: (Co1-342:6), 3-d wave function, probability, momentum representation

16. Consider the 3-dimensional wave function

Ψ(~r ) = A exp

[

−∑

i

|xi|/(2ai)

]

,

where the sum runs over the three Cartesian coordinates and the ai’s are real positive lengthparameters.

a) Calculate the normalization factor A. HINT: Recall that the integrand is |Ψ(~r )|2 =Ψ(~r )∗Ψ(~r ). I’m always forgetting this myself when the function is pure real and there isno imaginary part to remind me of it.

b) Calculate the probability that a measurement of xi will yield a result between 0 and ai,where i could be any of the three coordinates. HINT: There are no restrictions on valuesof the other coordinates: they could be anything at all. Thus one just integrates over allof those other coordinate positions remembering normalization of course.

c) Calculate the probability that simultaneous measurements of xj and xk will yield resultsin the ranges −aj to aj and −ak to ak, respectively. The j and k could be any pair of thetwo coordinates. HINT: Remember the hint for part (b).

d) Calculate the probability that a measurment of momentum will yield a result in the elementdpi dpj dpk centered at the point pi = pj = 0, pk = h−/ak. HINT: You will need to findthe momentum representation of the state.

Chapt. 3 Infinite Square Wells and Other Wells


003 qmult 00050 1 1 1 easy memory: infinite square well1. In quantum mechanics, the infinite square well can be regarded as the prototype of:

a) all bound systems. b) all unbound systems.c) both bound and unbound systems. d) neither bound nor unbound systems.e) Prometheus unbound.

003 qmult 00100 2 4 2 moderate deducto-memory: infinite square well BCs2. In the infinite square well problem, the wave function and its first spatial derivative are:

a) both continuous at the boundaries.b) continuous and discontinuous at the boundaries, respectively.c) both discontinuous at the boundaries.d) discontinuous and continuous at the boundaries, respectively.e) both infinite at the boundaries.

003 qmult 00300 1 1 3 easy memory: boundary conditions3. Meeting the boundary conditions of bound quantum mechanical systems imposes:

a) Heisenberg’s uncertainty principle. b) Schrodinger’s equation. c) quantization.d) a vector potential. e) a time-dependent potential.

003 qmult 00400 1 1 5 easy memory: continuum of unbound states4. At energies higher than the bound stationary states there:

a) are between one and several tens of unbound states.b) are only two unbound states. c) is a single unbound state. d) are no states.e) is a continuum of unbound states.

003 qmult 00500 1 4 2 easy deducto-memory: tunneling5. “Let’s play Jeopardy! For $100, the answer is: This effect occurs because wave functions can

extend (in an exponentially decreasing way albeit) into the classically forbidden region: i.e., theregion where a classical particle would have negative kinetic energy.”

What is , Alex?

a) stimulated radiative emission b) quantum mechanical tunneling c) quantizationd) symmetrization e) normalization

003 qmult 00600 2 1 2 moderate memory: benzene ring model6. A simple model of the outer electronic structure of a benzene molecule is a 1-dimensional infinite

square well with:

a) vanishing boundary conditions. b) periodic boundary conditions.c) aperiodic boundary conditions. d) no boundary conditions.e) incorrect boundary conditions.

24

Chapt. 3 Infinite Square Wells and Other Wells 25


003 qfull 00100 2 3 0 moderate math: infinite square well in 1-d1. You are given the time-independent Schrodinger equation

Hψ(x) =

[

− h−2

2m

∂2

∂x2+ V (x)

]

ψ(x) = Eψ(x)

and the infinite square well potential

V (x) =

0 , x ∈ [0, a];∞ otherwise.

a) What must the wave function be outside of the well (i.e., outside of the region [0, a]) in orderto satisfy the Schrodinger equation? Why?

b) What boundary conditions must the wave function satisfy? Why must it satisfy these boundaryconditions?

c) Reduce Schrodinger’s equation inside the well to an equation of the same form as theCLASSICAL simple harmonic oscillator differential equation with all the constants combinedinto a factor of −k2, where k is newly defined constant. What is k’s definition?

d) Solve for the general solution for a SINGLE k value, but don’t impose boundary conditionsor normalization yet. A solution by inspection is adequate. Why can’t we allow solutions withE ≤ 0? Think carefully: it’s not because k is imaginary when E < 0.

e) Use the boundary conditions to eliminate most of the solutions with E > 0 and to imposequantization on the allowed set of distinct solutions (i.e., on the allowed k values). Givethe general wave function with the boundary conditions imposed and give the quantizationrule for k in terms of a dimensionless quantum number n. Note that the multiplication of awave function by an arbitrary global phase factor eiφ (where φ is arbitrary) does not create aphysically distinct wave function (i.e., does not create a new wave function as recognized bynature.) (Note the orthogonality relation used in expanding general functions in eigenfunctionsalso does not distinguish eigenfunctions that differ by global phase factors either: i.e., it givesthe expansion coefficients only for distinct eigenfunctions. So the idea of distinct eigenfunctionsarises in pure mathematics as well as in physics.)

f) Normalize the solutions.

g) Determine the general formula for the eigenenergies in terms of the quantum number n.

003 qfull 00400 2 3 0 moderate math: moments of infinite square wellExtra keywords: (Gr-29:2.4)

2. Calculate 〈x〉, 〈x2〉, 〈p〉, 〈p2〉, σx, and σp for the 1-dimensional infinite square well with range[0, a]. Recall the general solution is

ψ =

√

2

asin(kx) =

√

2

asin(nπ

ax)

,

where n = 1, 2, 3, . . . . Also check that the Heisenberg uncertainty principle is satisfied.

003 qfull 00450 2 3 0 moderate math: infinite square well features

26 Chapt. 3 Infinite Square Wells and Other Wells

3. The one-dimensional infinite square well with a symmetric potential and width a is

V =

0 for |x| ≤ a/2;∞ for |x| > a/2.

The eigenstates for infinite square well are given by

ψn(x) =

√

2

a×

cos(kx) for n = 1, 3, 5 . . .;sin(kx) for n = 2, 4, 6 . . .,

whereka

2=nπ

2and k =

nπ

a.

The n is the quantum number for eigenstates. The eigenstates have been normalized and areguaranteed orthogonal by the mathematics of Hermitian operators of the which the Hamiltonianis one. A quantum number is a dimensionless index (usually integer or half-integer) that specifiesthe eigenstates and eigenvalues somehow. The eigen-energies are given by

En =h−2k2

2m=

h−2

2m

(π

a

)2

n2 .

a) Verify the normalization of eigenstates.

b) Determine 〈x〉 for the eigenstates.

c) Determine 〈pop〉 for the eigenstates. HINT: Recall

pop =h−i

∂

∂x.

d) Determine 〈p2op〉 and the momentum standard deviation σp for the eigenstates.

e) Determine 〈x2〉 and the position standard deviation σx in the large n limit. HINT: Assumex2 can be approximated constant over one complete cycle of the probability density ψ∗

nψn

f) Now for the boring part. Determine 〈x2〉 and the position standard deviation σx exactlynow. HINT: There probably are several different ways of doing this, but there seem to beno quick tricks to the answer. The indefinite integral

∫

x2 cos(bx) dx =x2

bsin(bx) +

2

b2x cos(bx) − 2

b3sin(bx)

might be helpful.

g) Verify that the Heisenberg uncertainty principle

∆x∆p = σxσp ≥ h−2

is satisfied for the infinite square well case.

003 qfull 00500 3 5 0 tough thinking: mixed infinite square well stationary statesExtra keywords: (Gr-29:2.6)

4. A particle is in a mixed state in a 1-dimensional infinite square well where the well spans [0, a]and the solutions are in the standard form of Gr-26. At time zero the state is

Ψ(x, 0) = A [ψ1(x) + ψ2(x)] ,


where ψ1(x) and ψ2(x) are the time-independent 1st and 2nd stationary states of the infinitesquare well.

a) Determine the normalization constant A. Remember the stationary states are orthonormal.Also is the normalization a constant with time? Prove this from the general time evolutionequation

d〈Q〉dt

=

⟨

∂Q

∂t

⟩

+1

h−〈i[H,Q]〉 .

b) Now write down Ψ(x, t). Give the argument for why it is the solution. As a simplficationin the solution use

ω1 =E1

h−=

h−2m

(π

a

)2

,

where E1 is the ground state energy of the infinite square well.

c) Write out |Ψ(x, t)|2 and simplify it so that it is clear that it is pure real. Make use Euler’sformula: eix = cosx + i sinx. What’s different about our mixed state from a stationarystate?

d) Determine 〈x〉 for the mixed state. Note that the solution is oscillatory. What is theangular frequency wq and amplitude of the oscillation. Why would you be wrong if youramplitude was greater than a/2.

e) Determine 〈p〉 for the mixed state. As Peter Lorre (playing Dr. Einstein—Herman Einstein,Heidelberg 1919) said in Arsenic and Old Lace “the quick way, Chonny.”

f) Determine 〈H〉 for the mixed state. How does it compare to E1 and E2?

g) Say a classical particle had kinetic energy equal to the energy 〈H〉 found in the part (f)answer. The particle is bounces back and forth between the walls of the infinite squarewell. What would its angular frequency be in terms of ωq and the angular frequency foundin the part (d) answer.

003 qfull 00600 2 5 0 moderate thinking: revival timeExtra keywords: Gr-85 The hints make it possible as new test problem.

5. The revival time is the minimum time period for a wave function to repeat (i.e., to cycle backto its original form), or slightly less restrictively, for the probability density to repeat.

a) Say we had a system with eigen-energies given by the formula

En = E1f(n) + E0 ,

where E0 is a zero-point energy, n is a quantum number that runs 1, 2, 3, . . . or0, 1, 2, 3, . . . , f(n) is a strictly increasing function that always has an integer value, andf(1) = 1. What is the revival time (in the probability density sense) for general wavefunction Ψ(x, t) for this system? HINT: The zero-point energy gives a time-dependentglobal phase factor for any expansion in the stationary states, and thus cancels out of theprobability density. Assume orthonormal energy-eigen states and recall

Ψ(x, t) =∑

n

cne−iωntψn(x) ,

where ωn = En/h− are the angular frequencies and ψn(x) are the energy-eigenstates. Theperiod for ωn is 2π/ωn.

b) The eigen-energies for the infinite square well and the simple harmonic oscillator are,respectively,

En =h−2

2m

(π

a

)2

n2 and En =

(

n+1

2

)

h−ωcl ,

28 Chapt. 3 Infinite Square Wells and Other Wells

where n = 1, 2, 3, . . . for the infintie square well and 0, 1, 2, 3, . . . for the simple harmonicoscillator, m is the particle mass, a is the well width, and ωcl simple harmonic oscillatorfrequency which enters the quantum mechanical description as parameter in the simpleharmonic oscillator potential. What are the revival times for general wave functions forthese two systems?

c) What are the classical revival times for a particle in a infinite square well system and in asimple harmonic oscillator system in terms of, respectively, energy and ωcl? The classicaltimes are just the oscillation periods for the particles. The particle in the infinite squarewell is assumed to be just bouncing between the walls without loss of kinetic energy.

d) For what classical energy E in units of E1 (i.e., E/E1) are the quantum mechanical andclassical revival times equal for the infinite square well? What is the relationship betweenthe quantum mechanical and classical revival times for the simple harmonic oscillator?

003 qfull 01000 3 5 0 tough thinking: 3-d infinite cubical wellExtra keywords: (Gr-124:4.2), separation of Schrodinger equation

6. Consider an infinite cubical well or particle-in-a-box system. The potential is

V (x, y, x) =

0, for x, y, and z in the range 0 to a;∞, otherwise.

The wave functions must be zero at the boundaries for an infinite well recall.

a) Solve for the stationary states from the 3-dimensional Schrodinger equation and find theirenergies in terms of quantum numbers nx, ny, and nz. HINTS: Separate the Schrodingerequation into x, y, and z components. Identify the sum of the separation constants as energyor, if you prefer, energy times a constant. Solve separately matching the boundary conditionsand then assemble the normalized TOTAL SOLUTION. Of course, all three dimensionsbehave the same so only one of them really needs to be done—which is NOT to say that eachone is a total solution all by itself.

b) Is there energy degeneracy? Why?

c) Determine the 6 lowest energies and their degeneracy? HINTS: A systematic approach wouldbe fix an nmax = max(nx, ny, nz) and count all energies and their degeneracies governed by thatnmax. One works one’s way up from nmax = 1 to as high as one needs to go to encompass the 6lowest energies. Each nmax governs the energies between n2

max + 2 and 3n2max (where we have

written the in dimensionless form). Note, e.g., that states described by (nx = 4, ny = 1, nz = 1),(nx = 1, ny = 4, nz = 1), and (nx = 1, ny = 1, nz = 4) are all distinct and degenerate.

003 qfull 01100 1 3 0 easy math: pi-states of a benzene ringExtra keywords: (Ha-323:2.1)

7. Imagine that we have 6 free electrons in 1-d circular system of radius r = 1.53 A. This system isa simple model of a benzene ring molecule (C6H6) of 6 carbon atoms each bonded to a hydrogen(Ke-153). The carbons are bonded by bonded by a single-double bond superposition. The freeelectron system on the benzene constitute the benzene pi-states.

a) Obtain expressions for the eigenstates, wavenumbers, and eigen-energies of the freeelectrons. Re-express the wavenumbers and energies in terms of Angstroms and

electronvolts. Note h−2/(2m) = 3.81 eV-A2 for electrons. Sketch the energy level diagram.

b) One electron per carbon lies in the circular state for a benzene ring: these are the πelectrons. Assuming that two electrons can be found in any state, what is the total energyof the ground state configurations? NOTE: Two electrons can be found in any statebecause there are two spin states they can be found in. Thus the Pauli exclusion principleis maintained: i.e., only one electron can be found in any single-particle state (e.g., Gr-180).


c) What is the energy difference in eV between the lowest empty level and highest occupiedlevel for the ground state configuration? This is the radiation absorption threshold. What isthe threshold line wavelength in microns? In what wavelength regime is this line? NOTE:The constant hc = 1.23984 eV-µm.

d) Now imagine we broke the benzene ring, but magically kept the length constant. Obtainexpressions for the eigenstates, wavenumbers, and eigen-energies of the free electrons. Re-express the wavenumbers and energies in terms of Angstroms and electronvolts. Sketch theenergy levels on the previous energy level diagram.

e) What is the ground state energy for the broken ring. What is the change in ground stateenergy from the unbroken ring. This change is a contribution to the energy required tobreak the ring or the energy of a resonant π bond.

f) I know we said that somewhere that quantum mechanical bound states always had to haveE > Vmin. But in the ring case we had Vmin = 0, and we have a state with E = 0. So whydo we have this paradox? Is the paradox possible in 2 dimensions or 3 dimensions?

Chapt. 4 The Simple Harmonic Oscillator (SHO)


004 qmult 00100 2 4 1 moderate deducto-memory: SHO eigen-energies1. “Let’s play Jeopardy! For $100, the answer is: h−ω.

a) What is the energy difference between adjacent simple harmonic ocsillator energy levels,Alex?

b) What is the energy difference between adjacent infinite square well energy levels, Alex?c) What is the energy difference between most adjacent infinite square well energy levels,

Alex?d) What is the energy difference between the first two simple harmonic ocsillator energy levels

ONLY, Alex?e) What is the bar where physicists hang out in Las Vegas, Alex?


004 qfull 00100 2 3 0 moderate math: SHO ground state analyzedExtra keywords: (Gr-19:1.14)

1. The simple harmonic oscillator (SHO) ground state is

Ψ0(x, t) = Ae−β2x2/2−iE0t/h− ,

where

E0 =h−ω2

and β =

√

mω

h−.

a) Verify that the wave function satisfies the full Schrodinger equation for the SHO. Recallthat the SHO potential is V (x) = (1/2)mω2x2.

b) Determine the normalization constant A.

c) Calculate the expectation values of x, x2, p, and p2.

d) Calculate σx and σp, and show that they satisfy the Heisenberg uncertainty principle.

004 qfull 00200 2 3 0 moderate thinking: SHO classically forbiddenExtra keywords: (Gr-43:2.15) classical turning points

2. What is the probability is of finding a particle in the ground state of a simple harmonic oscillatorpotential outside of the classically allowed region: i.e., beyond the classical turning points?HINT: You will have to use a table of the integrated Gaussian function.

004 qfull 00300 2 5 0 moderate thinking: mixed SHO stationary statesExtra keywords: (Gr-43:2.17)

30

Chapt. 4 The Simple Harmonic Oscillator (SHO) 31

3. A particle in a simple harmonic oscillator (SHO) potential has initial wave function

Ψ(x, 0) = A [ψ0 + ψ1] ,

where A is the normalization constant and the ψi are the standard form 0th and 1st SHOeigenstates. Recall the potential is

V (x) =1

2mω2x2 .

Note ω is just an angular frequency parameter of the potential and not NECESSARILYthe frequency of anything in particular. In the classical oscillator case ω is the frequency ofoscillation, of course.

a) Determine A assuming it is pure real as we are always free to do.

b) Write down Ψ(x, t). There is no need to express the ψi explicitly. Why must this Ψ(x, t)be the solution?

c) Determine |Ψ(x, t)|2 in simplified form. There should be a sinusoidal function of time inyour simplified form.

d) Determine 〈x〉. Note that 〈x〉 oscillates in time. What is its angular frequency andamplitude.

e) Determine 〈p〉 the quick way using the 1st formula of Ehrenfest’s theorem. Check that the2nd formula of Ehrenfest’s theorem holds.

004 qfull 01000 3 5 0 tough thinking: infinite square well/SHO hybridExtra keywords: (Mo-424:9.4)

4. Say you have the potential

V (x) =

∞ , x < 0;1

2mω2x2 x ≥ 0.

a) By reflecting on the nature of the potential AND on the boundary conditions, identifythe set of Schrodinger equation eigenfunctions satisfy this potential. Justify your answer.HINTS: Don’t try solving the Schrodinger equation directly, just use an already knownset of eigenfunctions to identify the new set. This shouldn’t take long.

b) What is the expression for the eigen-energies of your eigenfunctions?

c) What factor must multiply the already-known (and already normalized) eigenfunctionsyou used to construct the new set you found in part (a) in order to normalize the neweigenfunctions? HINT: Use the evenness or oddness (i.e., definite parity) of the already-known set.

d) Show that your new eigenfunctions are orthogonal. HINT: Use orthogonality and thedefinite parity of the already-known set.

e) Show that your eigenfunctions form a complete set given that the already-known setwas complete. HINTS: Remember completeness only requires that you can expand anysuitably well-behaved function (which means I think it has to be piecewise continuous (Ar-435) and square-integrable (CDL-99) satisfying the same boundary conditions as the setused in the expansion. You don’t have to be able to expand any function. Also, use thecompleteness of the already-known set.

004 qfull 01100 3 5 0 tough thinking: Hermite polynomials 1

32 Chapt. 4 The Simple Harmonic Oscillator (SHO)

5. The generating function method is a powerful method for obtaining the eigenfunctions of Sturm-Liouville Hermitian operators and some of their general properties. One can possibly obtainwith only moderately arduous labor some special values, the norm value, a general series formulafor the eigenfunctions, and recurrence relations for iteratively constructing the complete set ofeigenfunctions. The only problem is who the devil thought up the generating function?

In the case of Hermite polynomials, the generating function—which may or may not havebeen thought up by French mathematician Charles Hermite (1822–1901)—is

g(x, t) = e−t2+2tx =

∞∑

n=0

Hntn

n!

(Ar-609ff; WA-644). The Hn are the Hermite polynomials: they are functions of x and n istheir order.

Actually, the HERMITE EQUATION needs a weight function e−x2

to be put in Sturm-Liouville self-adjoint form (Ar-426, WA-486). Alternatively, the Hermite polynomials times

e−x2/2 satisfy a Sturm-Liouville Hermitian operator equation which happens to be the time-independent Schrodinger equation for the 1-dimensional quantum mechanical simple harmonicoscillator (Ar-612, WA-638). The 1-dimensional quantum mechanical simple harmonic oscillatoris one of those few quantum mechanical systems with an analytic solution.

NOTE: The parts of this question are independent: i.e., you should be able to do any ofthe parts without having done the other parts.

a) Find the 1st recurrence relation

Hn+1 = 2xHn − 2nHn−1

by differentiating both the generating function and its and series expansion with respect tot. This recurrence relation provides a means of finding any order of Hermite polynomial.HINT: You will need to re-index summations and make use of the uniqueness theorem ofpower series.

b) Find the 2nd recurrence relationH ′

n = 2nHn−1

by differentiating both the generating function and its and series expansion with respect tox. HINT: You will need to re-index summations and make use of the uniqueness theoremof power series.

c) Use the 1st recurrence relation to work out and tabulate the polynomials up to 3rd order:i.e., find H0, H1, H2, and H3. You can find the first two polynomials (i.e., the 0th and1st order polynomials) needed to start the recurrence process by a simple Taylor’s seriesexpansion of generating function.

d) Use the 1st recurrence relation to prove that the order of a Hermite polynomial agrees withits polynomial degree (which is the degree of its highest degree term) and that even orderHermite polynomials are even functions and the odd order ones are odd functions. Thelast result means that the Hermite polynomials have definite parity (i.e., are either even orodd functions). HINT: Use proof by induction and refer to collectively to the results tobe proven as “the results to be proven”. If you didn’t get H0 and H1 explicitly in part (c),you can assume H0 has degree 0 and H1 has degree 1.

004 qfull 01110 3 5 0 tough thinking: Hermite polynomials 26. Now for some more Hermite polynomial results. Recall the Hermite polynomial generating

function is

g(x, t) = e−t2+2tx =

∞∑

n=0

Hntn

n!


(Ar-609ff; WA-644). The Hn are the Hermite polynomials: they are functions of x and n istheir order. Also recall the two recurrence relations

Hn+1 = 2xHn − 2nHn−1 and H ′n = 2nHn−1

and the first few Hermite polynomials which are given in the table below.

Table: Hermite Polynomials

Order Polynomial

0 H0 = 11 H1 = 2x2 H2 = 4x2 − 23 H3 = 8x3 − 12x4 H4 = 16x4 − 48x2 + 125 H5 = 32x5 − 160x3 + 120x6 H6 = 64x6 − 480x4 + 720x2 − 120

NOTE: The parts of this question are largely independent: i.e., you should be able to domost parts without having done others.

a) Now for something challenging. Show that

g(x, t) = e−t2+2tx =

∞∑

ℓ=0

(−t2 + 2tx)ℓ

ℓ!=

∞∑

n=0

tn

n!

[n/2]∑

k=0

n!

(n− 2k)!k!(−1)k(2x)n−2k

which implies that

Hn =

[n/2]∑

k=0

n!

(n− 2k)!k!(−1)k(2x)n−2k .

Note

[n/2] =

n/2 for n even;(n− 1)/2 for n odd.

HINTS: You will have to expand (−t2 + 2tx)n in a binomial series and then re-order thesummation. A schematic table of the terms ordered in row by ℓ and in column by k makesthe re-ordering of the summation clearer. One should add up diagonals rather than rows.

b) Prove the following special results from the generating function:

H2n(0) = (−1)n (2n)!

n!, H2n+1(0) = 0 , Hn(x) = (−1)nHn(−x) .

The last results shows that the Hermite polynomials have definite parity: even for n even;odd for n odd.

c) What is called the Rodrigues’s formula for the Hermite polynomials can also be derivedfrom the generating function:

Hn = (−1)nex2 ∂n

∂xn(e−x2

) .

Derive this formula. HINTS: Write

g(x, t) = e−t2+2tx = ex2

e−(t−x)2

34 Chapt. 4 The Simple Harmonic Oscillator (SHO)

and note that∂f(t− x)

∂t= −∂f(t− x)

∂x.

d) Now show that Hermite’s differential equation

H ′′n − 2xH ′

n + 2nHn = 0

follows from the two recurrence relations. This result shows that the Hermite polynomialssatisfy Hermite’s differential equation.

e) Now consider the Hermite differential equation

h′′ − 2xh′ + 2νh = 0 ,

where ν is not necessarily an integer ≥ 0. Try a power series solution

h =

∞∑

ℓ=0

aℓxℓ ,

and show for sufficiently large ℓ and x that the series solutions approximate growingexponentials of the form ex2

and xex2

—unless ν is a positive or zero integer in whichcase one gets what kind of solution?

f) Hey this question is just going on and on. The Hermite differential equation cannot bewritten in an eigenproblem form with a Hermitian operator since the operator

∂2

∂x2− 2x

∂

∂x

is not, in fact, Hermitian. I won’t ask you to prove this since I don’t what to do that myselftonight. But if you substitute for Hn(x) (with n a positive or zero integer) the function

ψn(x)ex2/2

in the Hermite differential equation, you do an eigenproblem with a Hermitian operator.Find this eigenproblem equation. What are the eigenfunctions and eigenvalues? Arethe eigenfunctions square-integrable: i.e., normalizable in a wave function sense? Do theeigenfunctions have definite parity? Are the eigenvalues degenerate for square-integrablesolutions? Based on a property of eigenfunctions of a Hermitian operator what can yousay about the orthogonality of the eigenfunctions?

g) In order to normalize the eigenfunctions of part (i) in a wave function sense consider therelation

∞∑

m,n=0

sm

m!

tn

n!e−x2

HmHn = e−x2

g(x, s)g(x, t) = e−x2

e−s2+2sxe−t2+2tx .

Integrate both sides over all x and use uniqueness of power series to find the normalizationconstants and incidently verify orthogonality.

h) Now here in part infinity we will make the connection to physics. The simple harmonicoscillator time independent Schrodinger equation is

− h−2

2m

∂2ψ

∂y2+

1

2mω2y2ψ = Eψ .


One can reduce this to the dimensionless eigenproblem of part (i), by changing the variablewith

x = βy .

To find β, let

A =h−2

2mand B =

1

2mω2

and divide equation through by an unknown C, equate what needs to be equated, and solvefor C and β. What are the physical solutions and eigen-energies of the SHO eigenproblem?

i) Check to see if Charles Hermite (1822–1901) did think up the Hermite polynomialgenerating function at the St. Andrews MacTutor History of Mathematics Archive:

http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

HINTS: You don’t have to do this in a test mise en scene.

Chapt. 5 Free Particles and Momemtum Representation


005 qmult 00100 1 1 2 easy deducto-memory: definition free particle1. A free particle is:

a) bound. b) unbound. c) both bound and unbound.d) neither bound nor unbound. e) neither here nor there.

005 qmult 00200 1 4 5 easy deducto-mem: free particle system2. The free particle system is one with where the potential is:

a) the simple harmonic oscillator potential (SHO). b) a quasi-SHO potential.c) an infinite square well potential. d) a finite square well potential.e) zero (or a constant) everywhere.

005 qmult 00300 1 4 4 easy deducto-mem: free particle eigenfunction3. The general expression for the free particle energy eigenfunction in 1-dimension is:

a) eikx, where k = ±E. b) ekx, where k = ±E. c) ekx, where k = ±√

2mE/h−2.

d) eikx, where k = ±√

2mE/h−2. e) ekx2

, where k = ±√

2mE/h−2.

005 qmult 00400 1 4 1 easy deducto-mem: free particle normalization 14. The free particle energy eigenfunctions are not physical states that a particle can actually be

in because they:

a) can’t be normalized (i.e., they arn’t square-integrable).b) can be normalized (i.e., they are square-integrable).c) are growing exponentials.d) don’t exist.e) do exist.

005 qmult 00500 1 1 3 easy memory: free particle normalization 25. The free particle stationary states:

a) can be occupied by a particle. b) can be occupied by two particles.c) cannot actually be occupied by a particle. d) are unknown.e) are normalizable.


005 qfull 00100 2 5 0 easy thinking: momentum representationExtra keywords: (Gr-49:2.21)

1. The initial wave function of a free particle is

Ψ(x, 0) =

A , x ∈ [−a, a];0 , otherwise,

36

Chapt. 5 Free Particles and Momemtum Representation 37

where a and A are positive real numbers. The particle is in a completely zero potentialenvironment since it is a free particle.

a) Determine A from normalization.

b) Determine ψ(k) = Ψ(k, 0) the time-zero wavenumber representation of the particle state.It is the Fourier transform of Ψ(x, 0). What is Ψ(k, t)? Sketch ψ(k). Locate the globalmaximum and the zeros of ψ(k). Give the expression for the zeros (i.e., for the location ofthe zeros).

c) Determine the wavenumber space probability density |Ψ(k, t)|2 and show then that Ψ(k, t)is normalized in wavenumber space. (You can use a table integral.) Sketch |Ψ(k, t)|2 andlocate the global maximum and the zeros. Give the expression for the zeros.

d) Crudely estimate and then calculate exactly σx, σk, and σp for time zero. Are the resultsconsistent with the Heisenberg uncertainty principle?

005 qfull 00200 3 3 0 tough math: k-representation of half exponentialExtra keywords: (Mo-140:4.4)

2. At time zero, a wave function for a free particle in a zero-potential 1-dimensional space is:

Ψ(x, 0) = Ae−|x|/ℓeik0x .

a) Determine the normalization constant A. HINT: Remember it’s Ψ(x, 0)∗Ψ(x, 0) thatappears in the normalization equation.

b) Sketch the x-space probability density |Ψ(x, 0)|2. What is the e-folding distance of theprobability density?

NOTE: The e-folding distance is a newish term that means the distance in which anexponential function changes by a factor of e. It can be generalized to any function f(x)using the formula

xe =

∣

∣

∣

∣

f(x)

f(x)′

∣

∣

∣

∣

,

where xe is the generalized e-folding distance. The generalized e-folding distance is onlylocally valid to the region near the x where the functions are evaluated. The generalizede-folding distance is also sometimes called the scale height. If f(x) were an exponentialfunction, xe would be the e-folding distance in the narrow sense. If f(x) were a linearfunction, xe would be the distance to the zero of the function.

c) Show that the wavenumber representation of free particle state is

Ψ(k) =

√

2ℓ

π

1

1 + (k0 − k)2ℓ2.

This, of course, is the Fourier transform of Ψ(x, 0). Recall the wavenumber representationis time-independent since the wavenumber eigenstates are the stationary states of thepotential.

d) Confirm that Ψ(k) is normalized in wavenumber space. HINTS: You will probably need anintegral table—unless you’re very, very good. Also remember it’s Ψ(k)∗Ψ(k) that appears inthe normalization integral; always easy to forget this when dealing with pure real functions.

e) Write down the time-dependent solution Ψ(x, t) in Fourier transform form? Don’t try toevaluate the integral. What is ω in terms of k and energy E?

005 qfull 00300 3 3 0 tough math: Gaussian free wave packet spreading

38 Chapt. 5 Free Particles and Momemtum Representation

Extra keywords: (Gr-50:2.22)3. A free particle has an initial Gaussian wave function

Ψ(x, 0) = Ae−ax2

,

where a and A are real positive constants.

a) Normalize Ψ(x, 0). HINT: Recall that the integrand is |Ψ(x, t)|2 = Ψ(x, t)∗Ψ(x, t). I’malways forgetting this myself when the function is pure real and there is no imaginary partto remind me of it.

b) Determine the wavenumber representation Ψ(k) (which is time-independent). This involvesa Gaussian integral where you have to complete the square in an exponential exponent.Note

exp[−(ax2 + bx)] = exp

[

−a(

x2 +b

ax+

b2

4a2

)]

exp

(

b2

4a2

)

= exp

[

−a(

x+b

2a

)2]

exp

(

b2

4a

)

.

The exponetial factor exp(

b2/4a)

comes out of the integral and the integral over the wholex-axis is just a simple Gaussian integral.

c) Determine Ψ(x, t). You have to again do a Gaussian integral where you have to completethe square in an exponential exponent. It’s not that hard to do, but it is tedious and smallerrors can mess things up.

d) Find the probability density |Ψ(x, t)|2. This should be a Gaussian if all goes well. Sketchthe function and identify the standard deviation σ. What happens to the probabilitydensity with time. HINT: Note the identities

[

exp

(

a+ ib

c+ id

)]∗

=

exp

[

ac+ i(bc− ad) + bd

c2 + d2

]∗

=

exp

[

ac− i(bc− ad) + bd

c2 + d2

]

=

[

exp

(

a− ib

c− id

)]

and(√

a+ ib)∗

=(√

reiφ)∗

=(√

reiφ/2)∗

=√re−iφ/2 ,

where a+ ib magnitude and phase are r =√a2 + b2 and φ = tan−1(b/a), respectively.

e) Find 〈x〉, 〈x2〉, σx, 〈p〉, 〈p2〉, and σp. HINT: These results follow immediately from theGaussian nature of the functions in parts (d) and (b).

f) Check that the Heisenberg uncertainty principle is satisfied. Does the equality ever hold?What’s true about the wave function at the time when the equality holds that is not trueat other times?

005 qfull 00400 3 5 0 tough thinking: general free wave packet spreadingExtra keywords: (CDL-342:4)

4. Consider a free particle in one dimension.

a) Show using Ehrenfest’s theorem that 〈x〉 is linear in time and that 〈p〉 is constant.

Chapt. 5 Free Particles and Momemtum Representation 39

b) Write the equation of motion (time evolution equation) for 〈p2〉, then 〈[x, p]+〉 (the subscript+ indicates anticommutator), and then 〈x2〉: i.e., obtain expressions for the time derivativesof these quantities. Simplify the expressions for the derivatives as much as possible, butwithout loss of generality. You should get nice compact formal results. Integrate thesederivatives with respect to time and remember constants of integration.

c) Using the results obtained in parts (a) and (b) and for suitable choice of one of the constantsof integration, show that

(

∆x2)

(t) =1

m2

(

∆p2)

0t2 +

(

∆x2)

0

where(

∆x2)

0and

(

∆p2)

0are the initial standard deviations.

005 qfull 00500 2 5 0 moderate thinking: x-op and k-op in x and k representation5. For a free particle in one dimension, the position and wavenumber representations of the state

expanded in eigenfunctions of the other representation, are, respectively:

Ψ(x, t) =

∫ ∞

−∞

φ(k, t)eikx

√2π

dk =

∫ ∞

−∞

φ(k)ei(kx−ωt)

√2π

dk

and

Φ(k, t) = φ(k)e−iωt =

[∫ ∞

−∞

Ψ(x, 0)e−ikx

√2π

dx

]

e−iωt ,

where

ω =h−k2

2m.

a) The expectation value for a general observable in Q the position representation is

〈Q〉 =

∫ ∞

−∞

Ψ(x, t)∗QΨ(x, t) dx ,

where Q can depend on x and differentiations with respect to x, but not on k nordifferentiations with respect to k. Substitute for Ψ(x, t) using the wavenumber expansionand find an expression that is a triple integral.

b) The wavenumber observable in the position representation is

kop =1

h−pop =

1

i

∂

∂x,

where pop is the momentum observable in the position representation. Using the result ofthe part (a) answer, identify the wavenumber observable in the wavenumber representation.

c) The position observable xop in the position representation is just the coordinate x. Usingthe result of the part (a) answer, identify the position observable in the wavenumberrepresentation.

005 qfull 00510 3 5 0 tough thinking: x-op and p-op in x and p representationExtra keywords: (Gr-117:3.51) a very general solution is given

6. In the position representation, the position operator xop is just x, a multiplicative variable. Themomentum operator pop in the position representation is

pdif =h−i

∂

∂x.

40 Chapt. 5 Free Particles and Momemtum Representation

where we use the subsript “dif” here to indicate explicitly that this is a differentiating operator.

a) Find the momentum operator pop in the momentum representation. HINTS: Operatewith pdif on the Fourier transform expansion of a general wave function

Ψ(x, t) =

∫ ∞

−∞

Ψ(p, t)eipx/h−

√2πh−

dp

and work the components of the integral around (using whatever tricks you need) until youhave

pdifΨ(x, t) =

∫ ∞

−∞

f(p, t)eipx/h−

√2πh−

dp .

The function f(p, t) is the Fourier transform of pdifΨ(x, t) and operator acting on Ψ(p, t)to give you f(p, t) is the momentum operator in the momentum representation.

b) Find the position operator xop in the momentum representation. HINTS: The same asfor part (a), mutatis mutandis: find the Fourier transform of wave function xΨ(x, t), etc.

c) What are the momentum representation versions of xk and pℓdif?

d) What is the momentum representation versions of xkpℓdif and pℓ

difxk. By the by, you should

remember how to interpret xkpℓdif and pℓ

difxk: they are two successive operators acting on

understood function to the right. Explicitly for a general function f(x), one could write

xkpℓdiff(x) = xk

[

pℓdiff(x)

]

and

pℓdifx

kf(x) = pℓdif

[

xkf(x)]

,

Unfortunately, there is sometimes ambiguity in writing formulae with operators: try to beclear.

e) What is the momentum representation version of Q(x, pdif) where Q is any linearcombination of powers of x and pdif including mixed powers. HINTS: Consider the generalterm

. . . xkpℓdifx

mpndif

and figure out what commutes with what.

f) Show that the expectation value of Q is the same in both representations. HINT:Remember the Dirac delta function

δ(k − k′) =

∫ ∞

−∞

ei(k−k′)x

2πdx .

Chapt. 6 Foray into Advanced Classical Mechanics


006 qmult 00100 1 1 2 easy memory: Newton’s 2nd law1. Classical mechanics can be very briefly summarized by:

a) Newton’s Principia.b) Newton’s 2nd law.c) Lagrange’s Traite de mecanique analytique.d) Euler’s 80 volumes of mathematical works.e) Goldstein 3rd edition.

006 qmult 00200 1 4 1 easy deducto-memory: Lagrangian formulationExtra keywords: see GPS-12, 17, and 48

2. “Let’s play Jeopardy! For $100, the answer is: A formulation of classical mechanics that isusually restricted to systems with holonomic or semi-holonomic virtual-displacement worklessconstraints without dissipation and uses the function L = T − V .”

a) What is the Lagrangian formulation, Alex?b) What is the Hamiltonian formulation, Alex?c) What is the Leibundgutian formulation, Alex?d) What is the Harrisonian formulation, Alex?e) What is the Sergeant Schultzian formulation, Alex?

006 qmult 00300 1 1 5 easy memory: Hamilton’s principle3. A fruitful starting point for the derivation of Lagrange’s equations is:

a) Lagrange’s lemma.b) Newton’s scholium.c) Euler’s conjecture.d) Laplace’s hypothesise) Hamilton’s principle.


006 qfull 01000 2 5 0 moderate thinking: Lorentz force1. The Lorentz force

~F = q

(

~E +~v

c× ~B

)

(here expressed in Gaussian units: Ja-238) can be obtained from Lagrange’s equations using aLagrangian containing a generalized potential

U = q

(

φ− ~v

c· ~A)

,

41

42 Chapt. 6 Foray into Advanced Classical Mechanics

where φ is the electric potential and ~A is the vector potential of electromagnetism. TheLagrangian is L = T − U , where T is the kinetic energy. Lagrange’s equations are

d

dt

(

∂L

∂qi

)

− ∂L

∂qi= 0 ,

where qi is a generalized coordinate (not charge) and qi is the total time derivative of q: (i.e.,the rate of change of qi which describes an actual particle.

Work from the Lorentz force expression for component i to

Fi = − ∂U

∂xi+d

dt

(

∂U

∂xi

)

,

where the xi are the Cartesian coordinates of a particle (xi are the particle velocity components).Then verify that

mxi = Fi

follows from the Lagrange equations.

You may need some hints. Recall that

~E = −∇φ− 1

c

∂ ~A

∂tand B = ∇× ~A

(Ja-219). The Levi-Civita symbol εijk will be useful since

(∇×A)i = εijk∂

∂xjAk .

where Einstein summation has been used. Recall

εijk =

1, ijk cyclic;−1, ijk anticyclic;0, if two indices the same.

The identity (with Einstein summation)

εijkεiℓm = δjℓδkm − δjmδkℓ

is also useful. I’ve never found an elegant derivation of this last identity: the only proof seemsto be by exhaustion. Note also that the total time derivative is interpreted as the rate of changeof a quantity as the particle moves. Thus

dAi

dt=∂Ai

∂t+∂Ai

∂xj

dxj

dt=∂Ai

∂t+∂Ai

∂xjxj =

∂Ai

∂t+∂Ai

∂xjvj ,

where we again use Einstein summation.

Chapt. 7 Linear Algebra


007 qmult 00100 1 1 5 easy memory: vector addition1. The sum of two vectors belonging to a vector space is:

a) a scalar.b) another vector, but in a different vector space.c) a generalized cosine.d) the Schwarz inequality.e) another vector in the same vector space.

007 qmult 00200 1 4 4 easy deducto-memory: Schwarz inequality2. “Let’s play Jeopardy! For $100, the answer is: |〈α|β〉|2 ≤ 〈α|α〉〈β|β〉.”

What is , Alex?

a) the triangle inequality b) the Heisenberg uncertainty principlec) Fermat’s last theorem d) the Schwarz inequalitye) Schubert’s unfinished last symphony

007 qmult 00300 1 4 5 easy deducto-memory: Gram-Schmidt procedure3. Any set of linearly independent vectors can be orthonormalized by the:

a) Pound-Smith procedure. b) Li Po tao. c) Sobolev method.d) Sobolev-P method. e) Gram-Schmidt procedure.

007 qmult 00400 1 4 4 moderate memory: definition unitary matrix4. A unitary matrix is defined by the expression:

a) U = UT , where superscript T means transpose. b) U = U †. c) U = U∗.d) U−1 = U †. e) U−1 = U∗.

007 qmult 00500 2 3 4 moderate math: trivial eigenvalue problem5. What are the eigenvalues of

(

1 −ii 1

)

?

a) Both are 0. b) 0 and 1. c) 0 and −1. d) 0 and 2. e) −i and 1.

007 qmult 00600 1 4 5 moderate memory: riddle Hermitian matrix6. Holy peccant poets Batman, it’s the Riddler.

I charge to the right and hit on a ket,and if it’s not eigen, it’s still in the set,I charge to left and with a quick drawmake a new bra from out of a bra.

Not fish nor fowl nor quadratic,not uncanny tho oft Q-mechanic,

43

44 Chapt. 7 Linear Algebra

and transposed I’m just the right meif also complexicated as you can see.

My arrows down drawn from quivered,the same when sped to the world deliveredaside from a steady factor, rock of reality,mayhap of a quantum and that’s energy.

a) A unitary operator.b) A ket—no, no, a bra vector.c) An eigenvalue.d) Hamlet.e) A Hermitian matrix.


007 qfull 00090 1 5 0 easy thinking: ordinary vector spaceExtra keywords: (Gr-77:3.1)

1. Consider ordinary 3-dimensional vectors with complex components specified by a 3-tuple:(x, y, z). They constitute a 3-dimensional vector space. Are the following subsets of this vectorspace vector spaces? If so, what is their dimension? HINT: See Gr-435 for all the propertiesa vector space must have.

a) The subset of all vectors (x, y, 0).

b) The subset of all vectors (x, y, 1).

c) The subset of all vectors of the form (a, a, a), where a is any complex number.

007 qfull 00100 2 5 0 moderate thinking: vector space, polynomialExtra keywords: (Gr-78:3.2)

2. A vector space is constituted by a set of vectors |α〉, |β〉, |γ〉, . . . and a set of scalars a, b, c, . . .(ordinary complex numbers is all that quantum mechanics requires) subject to two operationsvector addition and scalar multiplication obeying the certain rules. Note it is the relationsbetween vectors that make them constitute a vector space. What they “are” we leave general.The rules are:

i) A sum of vectors is a vector:|α〉 + |β〉 = |γ〉 ,

where |α〉 and |β〉 are any vectors in the space and |γ〉 also in the space. Note we havenot defined what vector addition consists of. That definition goes beyond the generalrequirements.

ii) Vector addition is commutative:

|α〉 + |β〉 = |β〉 + |α〉 .

iii) Vector addition is associative:

(|α〉 + |β〉) + |γ〉 = |α〉 + (|β〉 + |γ〉) .

iv) There is a zero or null vector |0〉 such that

|α〉 + |0〉 = |α〉 ,

Chapt. 7 Linear Algebra 45

v) For every vector |α〉 there is an inverse vector | − α〉 such that

|α〉 + | − α〉 = |0〉 .

vi) Scalar multiplication of a vector gives a vector:

a|α〉 = |β〉 .

vii) Scalar multiplication is distributive on vector addition:

a(|α〉 + |β〉) = a|α〉 + a(|β〉) .

viii) Scalar multiplication is distributive on scalar addition:

(a+ b)|α〉 = a|α〉 + b|α〉 .

ix) Scalar multiplication is associative with respect to scalar multiplication:

(ab)|α〉 = a(b|α〉) .

x) One has0|α〉 = |0〉 .

xi) Finally, one has1|α〉 = |α〉 .

NOTE: Note that

|0〉 = 0|α〉 = [1 + (−1)]|α〉 = |α〉 + (−1)|α〉 ,

and thus we find that| − α〉 = −|α〉 .

So the subtraction of a vector is just the addition of its inverse. This is consistent with allordinary math.

If any vector in the space can be written as linear combination of a set of linearlyindependent vectors, that set is called a basis and is said to span the set. The number ofvectors in the basis is the dimension of the space. In general there will be infinitely many basesfor a space.

Finally the question. Consider the set of polynomials P (x) (with complex coefficients)and degree less than n. For each of the subsets of this set (specified below) answer the followingquestions: 1) Is the subset a vector space? Inspection usually suffices to answer this question.2) If not, what property does it lack? 3) If yes, what is the most obvious basis and what is thedimension of the space?

a) The subset that is the whole set.

b) The subset of even polynomials.

c) The subset where the highest term has coefficient a (i.e., the leading coefficient is a) and ais a general complex number, except a 6= 0.

d) The subset where P (x = g) = 0 where g is a general real number. (To be really clear, Imean the subset of polynomials that are equal to zero at the point x = g.)


e) The subset where P (x = g) = h where g is a general real number and h is a general complexnumber, except h 6= 0.

007 qfull 00110 2 5 0 moderate thinking: unique expansion in basisExtra keywords: (Gr-78:3.3)

3. Prove that the expansion of a vector in terms of some basis is unique: i.e., the set of expansioncoefficients for the vector is unique.

007 qfull 00200 3 5 0 tough thinking: Gram-Schmidt orthonormalizationExtra keywords: (Gr-79:3.4)

4. Say |αi〉 is a basis (i.e., a set of linearly independent vectors that span a vector space), butit is not orthonormal. The first step of the Gram-Schmidt orthogonalization procedure is tonormalize the (nominally) first vector to create a new first vector for a new orthonormal basis:

|α′1〉 =

|α1〉||α1||

,

where recall that the norm of a vector |α〉 is given by

||α|| = || |α1〉 || =√

〈α|α〉 .

The second step is create a new second vector that is orthogonal to the new first vector usingthe old second vector and the new first vector:

|α′2〉 =

|α2〉 − |α′1〉〈α′

1|α2〉|| |α2〉 − |α′

1〉〈α′1|α2〉 ||

.

Note we have subtracted the projection of |α2〉 on |α′1〉 from |α2〉 and normalized.

a) Write down the general step of the Gram-Schmidt procedure.

b) Why must an orthonormal set of non-null vectors be a linearly independent.

c) Is the result of a Gram-Schmidt procedure independent of the order the original vectorsare used? HINT: Say you first use vector |αa〉 of the old set in the procedure. The firstnew vector is just |αa〉 normalized: i.e., |α′

a〉=|αa〉/||αa||. All the other new vectors will beorthogonal to |α′

a〉. But what if you started with |αb〉 which in general is not orthogonalto |αa〉?

d) How many orthonormalized bases can an n dimensional space have in general? (Ignorethe strange n = 1 case.) HINT: Can’t the Gram-Schmidt procedure be started with anyvector at all in the vector space?

e) What happens in the procedure if the original vector set |αi〉 does not, in fact, consistof all linearly independent vectors? To understand this case analyze another apparentlydifferent case. In this other case you start the Gram-Schmidt procedure with n originalvectors. Along the way the procedure yields null vectors for the new basis. Nothing canbe done with the null vectors: they can’t be part of a basis or normalized. So you justput those null vectors and the vectors they were meant to replace aside and continue withthe procedure. Say you got m null vectors in the procedure and so ended up with n −mnon-null orthonormalized vectors. Are these n −m new vectors independent? How manyof the old vectors were used in constructing the new n−m non-null vectors and which oldvectors were they? Can all the old vectors be recontructed from the the new n−m non-nullvectors? Now answer the original question.

f) If the original set did consist of n linearly independent vectors, why must the neworthonormal set consist of n linearly independent vectors? HINT: Should be just acorollary of the part (e) answer.


g) Orthonormalize the 3-space basis consisting of

|α1〉 =

1 + i1i

, |α2〉 =

i31

, and |α3〉 =

0320

.

Input the vectors into the procedure in the reverse of their nominal order: why might amarker insist on this? Note setting kets equal to columns is a lousy notation, but you-allknow what I mean. The bras, of course, should be “equated” to the row vectors. HINT:Make sure you use the normalized new vectors in the construction procedure.

007 qfull 00300 2 3 0 moderate math: prove the Schwarz inequalityExtra keywords: (Gr-80:3.5)

5. As Andy Rooney says (or used to say if this problem has reached the stage where only old fogiesremember that king of the old fogies) don’t you just hate magic proofs where you start fromsome unmotivated expression and do a number of unmotivated steps to arrive at a result thatyou could never have been guessed from the way you were going about getting it. Well sans toomany absurd steps, let us see if we can prove the Schwarz inequality

|〈α|β〉|2 ≤ 〈α|α〉〈β|β〉

for general vectors |α〉 and |β〉. Note the equality only holds in two cases. First when |β〉 = a|α〉,where a is some complex constant. Second, when either or both of |α〉 and |β〉 are null vectors:in this case, one has zero equals zero.

NOTE: A few facts to remember about general vectors and inner products. Say |α〉 and |β〉are general vectors. By the definition of the inner product, we have that 〈α|β〉 = 〈β|α〉∗.This implies that 〈α|α〉 is pure real. If c is a general complex number, then the inner productof |α〉 and c|β〉 is 〈α|c|β〉 = c〈α|β〉. Next we note that that another inner-product propertyis that 〈α|α〉 ≥ 0 and the equality only holds if |α〉 is the null vector. The norm of |α〉 is||α|| =

√

〈α|α〉 and |α〉 can be normalized if it is not null: i.e., for |α〉 not null, the normalizedversion is |α〉 = |α〉/||α||.a) In doing the proof of the Schwarz inequality, it is convenient to have the result that the

bra c∗〈γ| = 〈γ|c∗ corresponds the ket c|γ〉, where |γ〉 is a general vector and c is a generalcomplex number. Prove this correspondance. HINT: Define |ǫ〉 = c|γ〉, take the innerproduct with general vector |δ〉, and do some manipulation making use of general vectorand inner-product properties.

b) The next thing to do is to figure out what the Schwarz inequality is saying about vectorsincluding those 3-dimensional things we have always called vectors. Let us a restrict thegenerality of |α〉 by demanding it not be a null vector for which the Schwarz inequalityis already proven. Since |α〉 is not null, it can be normalized. Let |α〉 = |α〉/||α|| be thenormalized version of |α〉. Divide the Schwarz inequality by ||α||2. Now note that thecomponent of |β〉 along the |α〉 direction is

|β‖〉 = |α〉〈α|β〉 .

Evaluate 〈β‖|β‖〉. Now what is the Schwarz inequality telling us.

c) The vector component of |β〉 that is orthogonal to |α〉 (and therefore |β‖〉) is

|β⊥〉 = |β〉 − |β‖〉 .

Prove this and then prove the Schwarz inquality itself (for |α〉 not null) by evaluating 〈β|β〉expanded in components. Schwarz inequality for |α〉 not a null vector. What if |α〉 is a nullvector?


007 qfull 00310 1 3 0 easy math: find a generalized angleExtra keywords: (Gr-80:3.6)

6. The general inner-product vector space definition of generalized angle according to Gr-440 is

cos θgen =|〈α|β〉|

√

〈α|α〉〈β|β〉,

where |α〉 and |β〉 are general non-zero vectors.

a) Is this definition completely consistent with the ordinary definition of an angle from theordinary vector dot product? Why or why not?

b) Find the generalized angle between vectors

|α〉 =

1 + i1i

and |β〉 =

4 − i0

2 − 2i

.

007 qfull 00400 1 3 0 easy math: prove triangle inequalityExtra keywords: (Gr-80:3.7)

7. Prove the triangle inequality:

||(|α〉 + |β〉)|| ≤ ||α|| + ||β|| .

HINT: Start with ||(|α〉 + |β〉)||2, expand, and use reality and the Schwarz inequality

|〈α|β〉|2 ≤ 〈α|α〉〈β|β〉 = ||α||2 × ||β||2 .

007 qfull 00500 3 3 0 tough math: simple matrix identitiesExtra keywords: (Gr-87:3.12)

8. Prove the following matrix identities:

a) (AB)T = BTAT, where superscript “T” means transpose.

b) (AB)† = B†A†, where superscript † means Hermitian conjugate.

c) (AB)−1 = B−1A−1.

d) (UV )−1 = (UV )† (i.e., UV is unitary) given that U and V are unitary. In other words,prove the product of unitary matrices is unitiary.

e) (AB)† = AB (i.e., AB is Hermitian) given that A and B are commuting Hermitianmatrices. Does the converse hold: i.e., does (AB)† = AB imply A and B are commutingHermitian matrices? HINTS: Find a trivial counterexample. Try B = A−1.

f) (A+B)† = A+B (i.e., A+B is Hermitian) given that A and B are Hermitian. Does theconverse hold? HINT: Find a trivial counterexample to the converse.

g) (U + V )† = (U + V )−1 (i.e., U + V is unitary) given that U and V are unitary—that is,prove this relation if it’s indeed true—if it’s not true, prove that it’s not true. HINT: Finda simple counterexample: e.g., two 2 × 2 unit matrices.

007 qfull 00510 2 5 0 moderate thinking: commuting operationsExtra keywords: (Gr-84)


9. There are 4 simple operations that can be done to a matrix: inversing, (−1), complexconjugating (∗), transposing (T ), and Hermitian conjugating (†). Prove that all these operationsmutually commute. Do this systematically: there are

(

4

2

)

=4!

2!(4 − 2)!= 6

combinations of the 2 operations. We assume the matrices have inverses for the proofs involvingthem.

007 qfull 00600 3 3 0 tough math: basis change resultsExtra keywords: (Gr-87:3.14)

10. Do the following.

a) Prove that matrix multiplication is preserved under similarity or linear basis change: i.e.,if AeBe = Ce in the e-basis, then AfBf = Cf in the f -basis where S is the basis changematrix from e-basis to the f -basis. Basis change does not in general preserve symmetry,reality, or Hermiticity. But since I don’t want to find the counterexamples, I won’t ask youto.

b) If He in the e-basis is a Hermitian matrix and the basis change to the f -basis U is unitary,prove that Hf is Hermitian: i.e., Hermiticity is is preserved.

c) Prove that basis orthonormality is preserved through a basis change U iff (if and only if)U is unitary.

007 qfull 00700 2 5 0 moderate thinking: square-integrable, inner productExtra keywords: no analog Griffiths’s problem, but discussion Gr-95–6, Gr2005-94–95

11. If f(x) and g(x) are square-integrable complex functions, then the inner product

〈f |g〉 =

∫ ∞

−∞

f∗g dx

exists: i.e., is convergent to a finite value. In other words, that f(x) are g(x) are square-integrable is sufficient for the inner product’s existence.

a) Prove the statement for the case where f(x) and g(x) are real functions. HINT: In doingthis it helps to define a function

h(x) =

f(x) where |f(x)| ≥ |g(x)| (which we call the f region);g(x) where |f(x)| < |g(x)| (which we call the g region),

and show that it must be square-integrable. Then “squeeze” 〈f |g〉.b) Now prove the statement for complex f(x) and g(x). HINTS: Rewrite the functions in

terms of their real and imaginary parts: i.e.,

f(x) = fRe(x) + ifIm(x)

and

g(x) = gRe(x) + igIm(x) .

Now expand

〈f |g〉 =

∫ ∞

−∞

f∗g dx

in the terms of the new real and imaginary parts and reduce the problem to the part (a)problem.


c) Now for the easy part. Prove the converse of the statement is false. HINT: Find sometrivial counterexample.

d) Now another easy part. Say you have a vector space of functions fi with inner productdefined by

∫ ∞

−∞

f∗j fk dx .

Prove the following two statements are equivalent: 1) the inner product property holds;2) the functions are square-integrable.

007 qfull 00800 2 3 0 moderate math: Gram-Schmidt, Legendre polynomialsExtra keywords: (Gr-96:3.25)

12. The Gram-Schmidt procedure can be used to construct a set of orthonormal vectors by linearcombination from a set of linearly independent, but non-orthonormal, vectors. It is a sort ofbrute force approach to use when more elegant methods of orthonormalization are not available.

The Gram-Schmidt procedure is as follows. Say |φn〉 are a set of N linearly independentvectors that are NOT assumed to be orthonormal. One has ordered them in some reasonablemanner indicated by the index n which increases from 1 to N . From this set we constructthe orthonormal set |un〉 of N normalized vectors where the hat symbol indicates theirnormalization. The unnormalized nth vector of the new set is given by

|un〉 = |φn〉 −n−1∑

ℓ=1

|uℓ〉〈uℓ|φn〉 ,

where the sum vanishes for n = 1, and the corresponding normalized vector is given by

|un〉 =|un〉||un||

,

where ||un|| =√

〈un|un〉 is the norm of |un〉.a) Prove the Gram-Schmidt procedure by induction.

b) In general there is an uncountable infinity of orthonormal sets that can be constructedfrom a non-orthonormal set. For example, consider ordinary vectors in 3-dimensionalEuclidean space. By rotation of a orthonormal set of unit vectors an uncountable infinity oforthonormal sets of unit vectors can be created. Different orthonormal sets can be createdusing the Gram-Schmidt procedure just by itself. Say that there are N vectors in linearlyindependent non-orthonomcal set and each vector is orthogonal with all the others in theset. Show that you can create at least N different orthogonal sets by the Gram-Schmidtprocedure.

c) Using the Gram-Schmidt procedure on the interval [−1, 1] (with weight function w = 1 inthe integral rule for the inner product) find the first three orthonormalized polynomialvectors starting from the linearly independent, but non-orthonomal set of polynomialvectors given by φn = xn for integer n ∈ [0,∞). Show that the orthonormalized setare the first three normalized Legendre polynomials Pn(x). The normalized Legendrepolynomials are given by

Pn(x) =

√

2n+ 1

2Pn(x) ,

where Pn(x) is a standard Legendre polynomial (e.g., Ar-547; WA-569). (Note an arbitraryphase factor eiθ can also be multiplied to a normalized vector.) The reason why theLegendre polynomials arn’t normalized is that the standard forms are what one getsstraight from the generating function. The generating function approach to the Legendrepolynomials allows you to prove many of their properties quickly (e.g., Ar-534, WA-553).


Table: Legendre Polynomials

Order n Pn

0 P0 = 1

1 P1 = x

2 P2 = (1/2)(3x2 − 1)

3 P3 = (1/2)(5x3 − 3x)

4 P4 = (1/8)(35x4 − 30x2 + 3)

5 P5 = (1/8)(63x5 − 70x3 + 15x)

Note—The polynomials are from Ar-541 and WA-554. The degree of a Legendre polynomialis given by its order number n (WA-557).

d) The normalized Legendre polynomials form a complete orthonormal basis for piecewisecontinuous, normalizable functions in the interval [−1, 1] (Ar-443). For a general polynomial ofdegree n, show that the expansion

Qn(x) =

∞∑

ℓ=0

Pℓ(x)〈Pℓ|Qn〉

actually truncates at ℓ = n. HINT: Consider xn and how could construct it from Legendrepolynomials starting with Pn(x).

e) The part (c) result suggests that if the Gram-Schmidt procedure is continued beyond the first3 vectors of the set φn, one will continue getting the normalized Legendre polynomials inorder. Prove this is so by comparing the Gram-Schmidt procedure result for 〈Pk|un〉 and theexpansion for |un〉 in the set |Pn〉. In evaluating 〈Pk|un〉, assume that the |uℓ〉 for ℓ < n are|Pℓ〉. HINT: The part (d) result is also needed for the proof.

007 qfull 00900 1 3 0 easy math: verifying a sinusoidal basisExtra keywords: (Gr-96:3.26)

13. Consider the set of trigonometric functions defined by

f(x) =

N∑

n=0

[an sin(nx) + bn cos(nx)]

on the interval [−π, π]. Show that the functions defined by

φk(x) =1√2πeikx , where k = 0,±1,±2, . . . ,±N

are an orthonormal basis for the trigonometric set. What is the dimension of the space spannedby the basis?

007 qfull 01000 2 3 0 moderate math: reduced SHO operator, HermiticityExtra keywords: (Gr-99:3.28), dimensionless simple harmonic oscillator Hamiltonian

14. Consider the operator

Q = − d2

dx2+ x2 .

a) Show that f(x) = e−x2/2 is an eigenfunction of Q and determine its eigenvalue.


b) Under what conditions, if any, is Q a Hermitian operator? HINTS: Recall

〈g|Q†|f〉∗ = 〈f |Q|g〉

is the defining relation for the Hermitian conjugate Q† of operator Q. You will have towrite the matrix element 〈f |Q|g〉 in the position representation and use integration by partsto find the conditions.

007 qfull 01100 2 5 0 moderate thinking: Hilbert space problemsExtra keywords: (Gr-103:3.33)

15. Do the following.

a) Show explicitly that any linear combination of two functions in the Hilbert space L2(a, b)is also in L2(a, b). (By explicitly, I mean don’t just refer to the definition of a vector spacewhich, of course requires the sum of any two vectors to be a vector.)

b) For what values of real number s is f(x) = |x|s in L2(−a, a)c) Show that f(x) = e−|x| is in L2 = L2(−∞,∞). Find the wavenumber space representation

of f(x): recall the wavenumber “orthonormal” basis states in the position representationare

〈x|k〉 =eikx

√2π

.

007 qfull 01200 2 5 0 moderate thinking: Hermitian conjugate of AB16. Some general operator and vector identities should be proven. Recall the definition of the

Hermitian conjugate of general operator Q is giveny by

〈α|Q|β〉 = 〈β|Q†|α〉∗ ,

where |α〉 and |β〉 are general vectors.

a) Prove that the bra corresponding to vector Q|β〉 is 〈β|Q† for general Q and |β〉. HINT:Define |β′〉 = Q|β〉 and then take the inner product of that vector with another generalvector |α〉 and use the definition of the Hermitian conjugate. I’d use bra 〈β′| on ket |α〉and then reverse the order complex conjugating.

b) Show that the Hermitian conjugate of a scalar c is just its complex conjugate.

c) Prove for operators, not matrices, that

(AB)† = B†A† .

The result is, of course, consistent with matrix representations of these operators. Butthere are representations in which the operators are not matrices: e.g., the momentumoperator in the position representation is differentiating operator

p =h−i

∂

∂x.

Our proof holds for such operators too since we’ve done the proof in the general operator-vector formalism.

d) Generalize the proof in part (c) for an operator product of any number.

e) Prove that (A+B)† = A† +B†.

f) Prove that c[A,B] is a Hermitian operator for Hermitian A and B only when c is pureimaginary constant.


007 qfull 01300 3 5 0 tough thinking: operators and matrices isomorphism17. Expressions involving vector linear transformations or operators often (always?) isomorphic

to the corresponding matrix expressions when the operators are represented by matrices inparticular orthonormal bases. We would like to demonstrate this statement for a few importantsimple cases. For clarity express operators with hats (e.g., A) and leave the correspondingmatrices unadorned (e.g., A). Consider a general orthonormal basis |i〉 where i serves as alabeling index. Recall that the unit operator using this basis is

I = |i〉〈i| ,

where we use the Einstein summation rule, and so there is a sum on i. Recall that the ijthmatrix element of A is defined by

Aij = 〈i|A|j〉 .

This definition means that the scalar product 〈a|A|b〉, where |a〉 and |b〉 are general vectors, canbe reexpressed by matrix expression:

〈a|A|b〉 = 〈a|i〉〈i|A|j〉〈j|b〉 = a∗iAijbj = ~a†A~b ,

where ~a and ~b are column vector n-tuples and where we have used the Einstein rule.

Prove that the following operator expressions are isomorphic to their corresponding matrixexpressions.

a) Sum of operators A+ B.

b) Product of operators AB.

c) Hermitian conjugation A†.

d) The identity (AB)† = B†A†.

007 qfull 01400 2 5 0 moderate thinking: bra ket projector completenessExtra keywords: (Gr-118:3.57) See also CDL-115, 138

18. For an inner product vector space there is some rule for calculating the inner product of twogeneral vectors: an inner product being a complex scalar. If |α〉 and |β〉 are general vectors,then their inner product is denoted by

〈α|β〉 ,

where in general the order is significant. Obviously different rules can be imagined for a vectorspace which would lead to different values for the inner products. But the rule must have threebasic properties:

〈β|α〉 = 〈α|β〉∗ ,(1)

〈α|α〉 ≥ 0 , where 〈α|α〉 = 0 if and only if |α〉 = |0〉,(2)

and

〈α|(

b|β〉 + c|γ〉)

= b〈α|β〉 + c〈α|γ〉 ,(3)

where |α〉, |β〉, and |γ〉 are general vectors of the vector space and b and c are general complexscalars.

There are some immediate corollaries of the properties. First, if 〈α|β〉 is pure real, then

〈β|α〉 = 〈α|β〉 .


Second, if 〈α|β〉 is pure imaginary, then

〈β|α〉 = −〈α|β〉 .

Third, if|δ〉 = b|β〉 + c|γ〉 ,

then〈δ|α〉∗ = 〈α|δ〉 = b〈α|β〉 + c〈α|γ〉

which implies〈δ|α〉 = b∗〈β|α〉 + c∗〈γ|α〉 .

This last result makes(

〈β|b∗ + 〈γ|c∗)

|α〉 = b∗〈β|α〉 + c∗〈γ|α〉

a meaningful expression. The 3rd rule for a vector product inner space and last corollarytogether mean that the distribution of inner product multiplication over addition happens inthe normal way one is used to.

Dirac had the happy idea of defining dual space vectors with the notation 〈α| for the dualvector of |α〉: 〈α| being called the bra vector or bra corresponding to |α〉, the ket vector or ket:“bra” and “ket” coming from “bracket.” Mathematically, the bra 〈α| is a linear function of thevectors. It has the property of acting on a general vector |β〉 and yielding a complex scalar: thescalar being exactly the inner product 〈α|β〉.

One immediate consequence of the bra definition can be drawn. Let |α〉, |β〉, and a begeneral and let

|α′〉 = a|α〉 .

Then〈α′|β〉 = 〈β|α′〉∗ = a∗〈β|α〉∗ = a∗〈α|β〉

implies that the bra corresponding to |α′〉 is given by

〈α′| = a∗〈α| = 〈α|a∗ .

The use of bra vectors is perhaps unnecessary, but they do allow some operations andproperties of inner product vector spaces to be written compactly and intelligibly. Let’s considera few nice uses.

a) The projection operator or projector on to unit vector |e〉 is defined by

Pop = |e〉〈e| .

This operator has the property of changing a vector into a new vector that is |e〉 timesa scalar. It is perfectly reasonable to call this new vector the component of the originalvector in the direction of |e〉: this definition of component agrees with our 3-dimensionalEuclidean definition of a vector component, and so is a sensible generalization of thatthe 3-dimensional Euclidean definition. This generalized component would also be thecontribution of a basis of which |e〉 is a member to the expansion of the original vector:again the usage of the word component is entirely reasonable. In symbols

Pop|α〉 = |e〉〈e|α〉 = a|e〉 ,

where a = 〈e|α〉.


Show that P 2op = Pop, and then that Pn

op = Pop, where n is any integer greater than orequal to 1. HINTS: Write out the operators explicitly and remember |e〉 is a unit vector.

b) Say we havePop|α〉 = a|α〉 ,

where Pop = |e〉〈e| is the projection operator on unit vector |e〉 and |α〉 is unknown non-nullvector. Solve for the TWO solutions for a. Then solve for the |α〉 vectors correspondingto these solutions. HINTS: Act on both sides of the equation with 〈e| to find an equationfor one a value. This equation won’t yield the 2nd a value—and that’s the hint for findingthe 2nd a value. Substitute the a values back into the original equation to determine thecorresponding |α〉 vectors. Note one a value has a vast degeneracy in general: i.e., manyvectors satisfy the original equation with that a value.

c) The Hermitian conjugate of an operator Q is written Q†. The definition of Q† is given bythe expression

〈β|Q†|α〉 = 〈α|Q|β〉∗ ,

where |α〉 and |β〉 are general vectors. Prove that the bra corresponding to ket Q|β〉 must〈β|Q† for general |α〉. HINTS: Let |β′〉 = Q|β〉 and substitute this for Q|β〉 in the definingequation of the Hermitian conjugate operator. Note operators are not matrices (althoughthey can be represented as matrices in particular bases), and so you are not free to usepurely matrix concepts: in particular the concepts of tranpose and complex conjugation ofoperators are not generally meaningful.

d) Say we define a particular operator Q by

Q = |φ〉〈ψ| ,

where |φ〉 and |ψ〉 are general vectors. Solve for Q†. Under what condition is

Q† = Q ?

When an operator equals its Hermitian conjugate, the operator is called Hermitian just asin the case of matrices.

e) Say |ei〉 is an orthonormal basis. Show that

|ei〉〈ei| = 1 ,

where we have used Einstein summation and 1 is the unit operator. HINT: Expand ageneral vector |α〉 in the basis.

Chapt. 8 Operators, Hermitian Operators, Braket Formalism


008 qmult 00090 1 4 5 easy deducto-memory: example Hilbert space1. “Let’s play Jeopardy! For $100, the answer is: A space of all square-integrable functions on thex interval (a, b).”

What is a , Alex?

a) non-inner product vector space b) non-vector space c) Dilbert spaced) Dogbert space e) Hilbert space

008 qmult 00100 1 1 3 easy memory: complex conjugate of scalar product2. The scalar product 〈f |g〉∗ in general equals:

a) 〈f |g〉. b) i〈f |g〉. c) 〈g|f〉. d) 〈f |i|g〉. e) 〈f |(−i)|g〉.

008 qmult 00200 1 4 3 easy deducto-memory: what operators do3. “Let’s play Jeopardy! For $100, the answer is: It changes a vector into another vector.”


a) wave function b) scalar product c) operator d) brae) telephone operator

008 qmult 00300 2 1 5 moderate memory: Hermitian conjugate of product4. Given general operators A and B, (AB)† equals:

a) AB. b) A†B†. c) A. d) B. e) B†A†.

008 qmult 00400 2 5 5 moderate thinking: general Hermitian conjugation5. The Hermitian conjugate of the operator λ|φ〉〈χ|ψ〉〈ℓ|A (with λ a scalar and A an operator) is:

a) λ|φ〉〈χ|ψ〉〈ℓ|A. b) λ|φ〉〈χ|ψ〉〈ℓ|A†. c) A|ℓ〉〈ψ|χ〉〈φ|λ∗ . d) A|ℓ〉〈ψ|χ〉〈φ|λ.e) A†|ℓ〉〈ψ|χ〉〈φ|λ∗.

008 qmult 00500 1 1 5 easy memory: compatible observables6. Compatible observables:

a) anticommute. b) are warm and cuddly with each other. c) have no hair.d) have no complete simultaneous orthonormal basis. e) commute.

008 qmult 00600 1 1 3 easy memory: parity operator7. The parity operator Π acting on f(x) gives:

df/dx. b) 1/f(x). c) f(−x). d) 0. e) a spherical harmonic.

008 qmult 00700 1 4 3 easy deducto-memory: braket expectation value8. Given the position representation for an expectation value

〈Q〉 =

∫ ∞

−∞

Ψ(x)∗QΨ(x) dx ,

56

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 57

what is the braket representation?

a) 〈Q|Ψ∗|Q〉. b) 〈Ψ∗|Q|Ψ〉. c) 〈Ψ|Q|Ψ〉. d) 〈Ψ|Q†|Ψ〉. e) 〈Q|Ψ|Q〉.

008 qmult 00800 1 4 3 easy deducto-memory: Hermitian eigenproblem9. What are the three main properties of the solutions to a Hermitian operator eigenproblem?

a) (i) The eigenvalues are pure IMAGINARY. (ii) The eigenvectors are guaranteedorthogonal, except for those governed by degenerate eigenvalues and these can always beorthogonalized. (iii) The eigenvectors DO NOT span all space.

b) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, exceptfor those governed by degenerate eigenvalues and these can always be orthogonalized.(iii) The eigenvectors span all space in ALL cases.

c) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, exceptfor those governed by degenerate eigenvalues and these can always be orthogonalized.(iii) The eigenvectors span all space for ALL FINITE-DIMENSIONAL spaces. Ininfinite dimensional cases they may or may not span all space. It is quantum mechanicspostulate that the eigenvectors of an observable (which is a Hermitian operator) span allspace.

d) (i) The eigenvalues are pure IMAGINARY. (ii) The eigenvectors are guaranteedorthogonal, except for those governed by degenerate eigenvalues and these can always beorthogonalized. (iii) The eigenvectors span all space in ALL FINITE-DIMENSIONALspaces. In infinite dimensional cases they may or may not span all space.

e) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, exceptfor those governed by degenerate eigenvalues and these can always be orthogonalized.

008 qmult 00900 1 4 5 easy deducto-memory: definition observable10. “Let’s play Jeopardy! For $100, the answer is: A physically significant Hermitian operator

possessing a complete set of eigenvectors.”


a) conjugate b) bra c) ket d) inobservable e) observable

008 qmult 01000 1 4 4 easy deducto-memory: time-energy inequality11. In the precisely-formulated time-energy inequality, the ∆t is:

a) the standard deviation of time.b) the standard deviation of energy.c) a Hermitian operator.d) the characteristic time for an observable’s value to change by one standard deviation.e) the characteristic time for the system to do nothing.

008 qmult 02000 1 1 5 easy memory: common eigensetsExtra keywords: See CDL-140ff

12. The statements “two observables commute” and “a common eigenset can be constructed fortwo observables” are

in flat contradiction. b) unrelated. c) in non-intersecting Venn diagrams.d) irrelevant in relation to each other. e) are equivalent in the sense that one impliesthe other.


008 qfull 00010 1 1 0 easy memory: what is a ket?

58 Chapt. 8 Operators, Hermitian Operators, Braket Formalism

1. What is a ket (representative general symbol |Ψ〉)?008 qfull 00015 1 1 0 easy memory: what is a bra?

2. What is a bra? (Representative general symbol 〈Ψ|.)008 qfull 00020 1 1 0 easy memory: why the braket formalism?

3. Why is quantum mechanics at the advanced level formulated in the braket formalism?

008 qfull 00030 2 5 0 moderate thinking: Hermiticity and expectation valuesExtra keywords: (Gr-94:3.21)

4. Recall the definition of Hermitian conjugate for a general operator Q is

〈α|Q|β〉 = 〈β|Q†|α〉∗ ,

where |α〉 and |β〉 are general vectors. If Q is Hermitian,

Q† = Q :

i.e., Q is its own Hermitian conjugate.

a) If Q is Hermitian, prove that the expectation value of a general vector |γ〉,

〈γ|Q|γ〉 ,

is pure real.

b) If the expectation value〈γ|Q|γ〉

is always pure real for general |γ〉, prove that Q is Hermitian. The statement to be proven isthe converse of the statement in part (a). HINT: First show that

〈γ|Q|γ〉 = 〈γ|Q†|γ〉 .

Then let |α〉 and |β〉 be general vectors and construct a vector |ξ〉 = |α〉 + c|β〉, where c is ageneral complex scalar. Note that the bra corresponding to c|β〉 is c∗〈β|. Expand both sides of

〈ξ|Q|ξ〉 = 〈ξ|Q†|ξ〉 ,

and then keep simplifying both sides making use of the of the first thing proven and the definitionof a Hermitian conjugate. Note that the It may be useful to note that

A†)† = A and (A+B)† = A† +B† ,

where A and B are general operators and You should be able to construct an expression wherechoosing c = 1 and then c = i requires Q = Q†.

c) What simple statement follows from the proofs in parts (a) and (b)?

008 qfull 00032 2 5 0 moderate thinking: energy operator5. We usually think of the Hamiltonian as being the “energy” operator, but term energy operator

is also used for the operator

Eop = ih− ∂

∂t

(Mo-184).

a) Find general expressions for the eigenvalues and normalizable eigenstates of Eop.


b) Show that Eop is not, in fact, a Hermitian operator in the space of physical wave functions.HINT: Recall the Hermitian conjugate of an operator Q is defined by

〈α|Q†|β〉 = 〈β|Q|α〉∗

(Gr-92), where Q† is the Hermitian conjugate and |α〉 and |β〉 are general vectors.

c) Is Eop formally a quantum mechanical observable? Does the projection postulate (Gr-105)on the measurement of energy apply to Eop? HINT: The last question is trickier than itseems.

008 qfull 00040 2 3 0 moderate math: solving an eigenproblemExtra keywords: (Gr-94:3.22) also diagonalizing a matrix.

6. Consider

T =

(

1 1 − i1 + i 0

)

.

a) Is T Hermitian?

b) Solve for the eigenvalues. Are they real?

c) Determine the normalized eigenvectors. Since eigenvectors are not unique to within a phasefactor, the marker insists that you arrange your eigenvectors so that the first componentof each is 1. Are they orthogonal?

d) Using the eigenvectors as columns construct the inverse of a unitary matrix transformationwhich applied to T gives a diagonalized version of T . Find this diagonalized version T d.What is special about the diagonal elements?

e) Compare the determinant det|T |, trace Tr(T ), and eigenvalues of T to those of T d.

008 qfull 00500 2 5 0 moderate thinking: x-op in general formalismExtra keywords: and k-op and p-op in general formalism too

7. The general formalism of quantum mechanics requires states to be vectors in Hilbert spaces anddynamical variables to be governed or determined (choose your verb) by observables (Hermitianoperators with complete sets of eigenstates: i.e., sets that form a basis for the Hilbert space).These requirements are a Procrustian bed for the position, wavenumber (or momentum), andkinetic energy operators. These operators have complete sets of eigenvectors in a sense, butthose eigenvectors arn’t in any Hilbert space, because they can’t be normalized. Neverthelesseverything works out consistently if some identifications are made. The momentum and kineticenergy eigenstates are the same as the wavenumber eigenstates, and so we won’t worry aboutthem. The momentum and kinetic energy eigenvalues are different, of course.

NOTE: Procrustes (he who stretches) was a robber (or cannibal) with a remarkable bed thatfit all guests—by racking or hacking according to whether small or tall. Theseus fit Procrustesto his own bed—and this was before that unfortunate incident with the Minotaur.

a) Consider the xop eigenproblem in the general form

xop|x〉 = x|x〉 ,

where x is the eigenvalue and |x〉 is the eigenvector. The eigenvalues x and eigenvectors|x〉’s form continuous, not discrete, sets. The unity operator for the xop basis is therefore

1 =

∫

dx |x〉〈x| ,

where it is implied that the integral is over all space. An ideal measurement of positionyields x and, by quantum mechanical postulate, puts the system is in state |x〉. But the


system can’t be really be in an unnormalizable state which is what the |x〉’s turn out tobe. The system can be in an integral linear combination of such states.

Expand a general state |Ψ〉 in the xop basis and identify what |Ψ〉 is in the positionrepresentation. Then identify what the inner product of two xop eigenvectors 〈x′|x〉 mustbe. Why can’t the |x〉 be in the Hilbert space? What is the position representation of |x〉?Prove that xop in the position operator is just x itself.

b) Repeat part (a), mutatis mutandis, for kop.

c) What must 〈x|k〉 be? This is just an identification, not a proof—there are no proofs.HINT: Expand |Ψ〉 in the wavenumber representation and then operate on |Ψ〉 with 〈x|.

d) What is 〈k|k′〉 if we insert the position representation unit operator given the answer topart (c).

e) In order to have consistency with past work what must the matrix elements 〈x|kop|x〉,〈x|Hop|x〉, and 〈k|xop|k〉 be. Note these are just identifications, not proofs—there areno proofs. We omit 〈k|Hop|k〉—you’re not ready for 〈k|Hop|k〉 as Jack Nicholson wouldsnarl—if he were teaching intro quantum.

008 qfull 00100 2 5 0 moderate thinking: expectation values two waysExtra keywords: (Gr-108:3.35)

8. Consider the observable Q and the general NORMALIZED vector |Ψ〉. By quantummechanics postulate, the expectation of Qn, where n ≥ 0 is some integer, for |Ψ〉 is

〈Qn〉 = 〈Ψ|Qn|Ψ〉 .

a) Assume Q has a discrete spectrum of eigenvalues qi and orthonormal eigenvectors |qi〉. Itfollows from the general probabilistic interpretation postulate of quantum mechanics, thatexpectation value of Qn for |Ψ〉 is given by

〈Qn〉 =∑

i

qni |〈qi|Ψ〉|2 .

Show that this expression for 〈Qn〉 also follows from the one in the preamble. What is∑

i |〈qi|Ψ〉|2 equal to?

b) Assume Q has a continuous spectrum of eigenvalues q and Dirac-orthonormal eigenvectors|q〉. (Dirac-orthonormal means that 〈q′|q〉 = δ(q′ − q), where δ(q′ − q) is the Dirac deltafunction. The term Dirac-orthonormal is all my own invention: it needed to be.) Itfollows from the general probabilistic interpretation postulate of quantum mechanics, thatexpectation value of Qn for |Ψ〉 is given by

〈Qn〉 =

∫

dq qn|〈q|Ψ〉|2 .

Show that this expression for 〈Qn〉 also follows from the one in the preamble. What is∫

dq |〈q|Ψ〉|2 equal to?

008 qfull 00200 2 5 0 moderate thinking: simple commutator identities9. Prove the following commutator identities.

a) [A,B] = −[B,A].

b)

∑

i

aiAi,∑

j

bjBj

=∑

ij

aibj [Ai, Bj ], where the ai’s and bj ’s are just complex numbers.


c) [A,BC] = [A,B]C +B[A,C].

d) [A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0. This has always seemed to me to be perfectlyuseless however true.

e) (c[A,B])† = c∗[B†, A†], where c is a complex number.

f) The special case of the part (e) identity when A and B are Hermitian and c is pureimaginary. Is the operator in this special case Hermitian or anti-Hermitian?

008 qfull 00300 3 5 0 tough thinking: nontrivial commutator identitiesExtra keywords: (Gr-111:3.41) but considerably extended.

10. Prove the following somewhat more difficult commutator identities.

a) Given[B, [A,B]] = 0 , prove [A,F (B)] = [A,B]F ′(B) ,

where A and B are general operators aside from the given condition and F (B) is a generaloperator function of B. HINTS: Proof by induction is probably best. Recall that any functionof an operator is (or is that should be) expandable in a power series of the operator: i.e.,

F (B) =

∞∑

n=0

fnBn ,

where fn are constants.

b) [x, p] = ih−.

c) [x, pn] = ih−npn−1. HINT: Recall the part (a) answer.

d) [p, xn] = −ih−nxn−1. HINT: Recall the part (a) answer.

008 qfull 01300 2 5 0 moderate thinking: general uncertainty principleExtra keywords: Gr-108’s proof

11. In quantum mechanics we believe that states are described by vectors in an abstract Hilbertspace. The Hilbert space is an inner product vector space among other things.

An inner product vector space (as physicists view it anyway) has vectors written |α〉 (thekets) and dual space vectors written 〈α| (the bras). We require that there is some rule suchthat the inner product operation

〈α|β〉yields a complex number for general vectors |α〉 and |β〉. The other properties required are:

〈β|α〉 = 〈α|β〉∗ ,

〈α|α〉 ≥ 0 ,

where the equality holds if and only if |α〉 = 0 (i.e., |α〉 is the null vector), and

〈α|(b|β〉 + c|γ〉) = b〈α|β〉 + c〈αγ〉 .

The norm of a vector |α〉 is defined by

||α|| =√

〈α|α〉 .

The vectors are transformed into new vectors by operators: e.g.,

|α′〉 = Q|α〉 ,


where Q is some operator. The Schwarz inequality

〈α|α〉〈β|β〉 ≥ |〈α|β〉|2

is a feature of inner product vector spaces. A concrete interpretation of the Schwarz inequalityis that a vector must have a magnitude larger than or equal to any component of the vectoralong any axis. This can be seen by dividing the Schwarz inequality equality by

〈α|α〉

assuming that |α〉 is not null and interpreting

〈α|β〉

as the component of |β〉 in the |α〉 direction where

|α〉 =|α〉||α|| .

In quantum mechanics, the inner product operation in the 1-dimensional positionrepresentation, for example, is defined by

〈α|β〉 =

∫ ∞

−∞

α(x)∗β(x) dx ,

where α(x) and β(x) are square integrable functions. The dynamical variables of quantummechanics are governed by physically relevant Hermitian operators with complete sets ofeigenvectors. These operators, not too cogently, are called observables. The mean value orexpectation value for an ensemble of identical states |α〉 of a dynamical variable governed byan observable Q is given by

〈α|Q|α〉 .

The expectation value can be viewed as the overlap integral between the states |α〉 and Q|α〉.The Hermitian conjugate Q† of an operator Q is defined by

〈α|Q|β〉 = 〈β|Q†|α〉∗ .

If Q† = Q, then Q is a Hermitian operator. Hermitian operators have only real expectationvalues. This follows from the general definition of Hermitian conjugation: if Q† = Q, then

〈α|Q|α〉 = 〈α|Q|α〉∗ .

This result implies that the eigenvalues of a Hermitian operator are pure real since they are theexpectation values of the eigenstates. We will not prove it here, but it is true also that non-degenerate eigenstates of a Hermitian operator are orthogonal and that the set of eigenstates ofa Hermitian operator are a complete set guaranteed if the space is finite dimensional (Gr-93).But if the space is infinite dimensional completeness is not guaranteed for the set of eigenstates(Gr-99). The Hermitian operators with complete sets can be quantum mechanical observables.

Given the foregoing (and anything I’ve left out), the general uncertainty principle is anecessary consequence. This principle is

σAσB ≥ 1

2|〈i[A,B]〉| ,


where A and B are observables, [A,B] = AB−BA is the commutator of A and B, and σA andσB are standard deviations for observables A and B. The standard deviation squared (i.e., thevariance) of A for a state |α〉, for example, is given by

σ2A = 〈α|(A − 〈A〉)2|α〉 ,

where 〈A〉 = 〈α|A|α〉. Since A is a Hermitian operator, one can write

σ2A = 〈α|(A − 〈A〉)(A − 〈A〉)|α〉

= 〈α|(A − 〈A〉)|f〉= 〈f |(A− 〈A〉)|α〉= 〈f |f〉 ,

where we have used the fact that expectation value of a Hermitian operator is real and havedefined |f〉 = (A− 〈A〉)|α〉.a) Define |g〉 = (B − 〈B〉)|α〉. Now prove that

σ2Aσ

2B ≥ |〈f |g〉2 .

b) Evaluate 〈f |g〉 in terms of expectation values of A and B for state |α〉. For brevity thestate can be left implied where convenient: i.e., 〈A〉 stands for 〈α|A|α〉, etc. Remember Aand B are Hermitian and 〈f |g〉 is in general complex.

c) Show that for any complex number z

|z|2 = [Re(z)]2 + [Im(z)]2 ,

where

Re(z) =1

2(z + z∗) and Im(z) =

1

2i(z − z∗) .

Then evaluateRe(〈f |g〉) and Im(〈f |g〉) .

d) In the classical analog case, A and B would just be dynamical variables whose statisticalproperties can be evaluated for some probability density. What does Re(〈f |g〉) correspondto classically?

e) Now show that the general uncertainty inequality follows from the inequality you found inthe part (a) answer. Is there a classical analog to the general uncertainty relation? Whyor why not?

f) Given that A and B are Hermitian, show that [A,B] is not Hermitian in general, but thati[A,B] is.

g) Find the uncertainty relation for operators x and p: i.e., the Heisenberg uncertaintyrelation.

h) What relationship between states |f〉 and |g〉 will make the general uncertainty an equality?

i) Apply the relationship for equality to the case of x and p and obtain a differential equationfor a one-dimensional state for which the Heisenberg uncertainty equality holds. Solve thisdifferential equation for the minimum uncertainty wave packet. What kind of wave packetdo you have? In what physical cases is the minimum uncertainty wave packet a solutionto the Schrodinger equation.


j) Explain why the Heisenberg uncertainty equality is useful for understanding the groundstates of many systems.

008 qfull 01400 3 5 0 tough thinking: general uncertainty principleExtra keywords: A dumb proof, and outdated by my current understanding.

12. You have a strange looking operator:

ℓ =δQ

∆Q+ i

δR

∆R.

where Q and R are general Hermitian operators, ∆Q and ∆R are the standard deviations of Qand R, and

δQ ≡ Q− 〈Q〉 and δR ≡ R− 〈R〉 .(a) Write down the Hermitian conjugate ℓ†.

(b) Show ℓ†ℓ a Hermitian operator and that it is a positive definite operator: i.e, that〈ℓ†ℓ〉 ≥ 0 . HINT: If you have to think about these results for more than a few seconds,then just assume them and go on.

(c) Multiply out ℓ†ℓ and gather the cross terms into a commutator operator. Substitutefor δQ and δR in the commutator using their definitions and simplify it.

(d) Evaluate the expectation value of the multiplied out ℓ†ℓ operator. Simplify itremembering the definition of standard deviation.

(e) Remembering the positive definite result from part (b), find an inequality satisfied by∆Q∆R.

(f) Since the whole of the foregoing mysterious procedure could have been done with Q andR interchanged in the definition of ℓ, what second inequality must be satisfied by ∆Q∆R.

(g) What third ∆Q∆R inequality is implied by two previous ones.

008 qfull 01500 2 3 0 moderate math: x-H uncertainty relationExtra keywords: (Gr-110:3.39)

13. Answer the following questions.

a) What is the uncertainty relation for operators x and H? Work it out until the expectationvalue is for the momentum operator p.

b) What is the time-dependent expression for any observable expectation value 〈Q〉 =〈Ψ(t)|Q|Ψ(t)〉 when the state |Ψ(t)〉 is expanded in the discrete set of stationary states(i.e., energy eigenstates) with their time-dependent factors included to allow for the timedependence of |Ψ(t)〉. Let the set of stationary states with explicit time dependence be

e−iEjt/h−|φj〉. Note functions of t commute with observables: observables may dependon time, but they don’t contain time derivatives.

c) If state |Ψ(t)〉 from part (b) is itself the stationary state e−iEjt/h−|φj〉, what the expectationvalue? Is the expectation value time independent?

d) Derive the special form of the uncertainty relation for operators x and H for the case ofa stationary state of H? What, in fact, is σH for a stationary state? HINT: RememberEhrenfest’s theorem.

008 qfull 01600 3 5 0 tough thinking: neutrino oscillationExtra keywords: (Gr-120:3.58)

14. There are systems that exist apart from 3-dimensional Euclidean space: they are internal degreesof freedom such intrinsic spin of an electron or the proton-neutron identity of a nucleon (isospin:


see, e.g., En-162 or Ga-429). Consider such an internal system for which we can only detecttwo states:

|+〉 =

(

10

)

and |−〉 =

(

01

)

.

This internal system is 2-dimensional in the abstract vector sense of dimensional: i.e., it can bedescribed completely by an orthonormal basis of consisting of the 2 vectors we have just given.When we measure this system we force it into one or other of these states: i.e., we make thefundamental perturbation. But the system can exist in a general state of course:

|Ψ(t)〉 = c+(t)|+〉 + c−(t)|−〉 =

(

c+(t)c−(t)

)

.

a) Given that |Ψ(t)〉 is NORMALIZED, what equation must the coefficients c+(t) and c−(t)satisfy.

b) For reasons only known to Mother Nature, the states we can measure (eigenvectors ofwhatever operator they may be) |+〉 and |−〉 are NOT eigenstates of the Hamiltonian thatgoverns the time evolution of internal system. Let the Hamiltonian’s eigenstates (i.e., thestationary states) be |+′〉 and |−′〉: i.e.,

H |+′〉 = E+|+′〉 and H |−′〉 = E−|−′〉 ,

where E+ and E− are the eigen-energies. Verify that the general state |Ψ(t)〉 expanded inthese energy eigenstates,

|Ψ(t)〉 = c+e−iE+t/h−|+′〉 + c−e

−iE−t/h−|−′〉

satisfies the general vector form of the Schrodinger equation:

ih− ∂

∂t|Ψ(t)〉 = H |Ψ(t)〉 .

HINT: This requires a one-line answer.

c) The Hamiltonian for this internal system has no differential operator form since there is nowave function. The matrix form in the |+〉 and |−〉 representation is

H =

(

f gg f

)

.

Given that H is Hermitian, prove that f and g must be real.

d) Solve for the eigenvalues (i.e., eigen-energies) of Hamiltonian H and for its normalizedeigenvectors |+′〉 and |−′〉 in column vector form.

e) Given at t = 0 that

|Ψ(0)〉 =

(

ab

)

show that

|Ψ(t)〉 =1√2(a+ b)e−i(f+g)t/h−|+′〉 +

1√2(a− b)e−i(f−g)t/h−|−′〉

and then show that

|Ψ(t)〉 = e−ift/h−[

a

(

cos(gt/h−)−i sin(gt/h−)

)

+ b

(

−i sin(gt/h−)cos(gt/h−)

)]

.


HINT: Recall the time-zero coefficients of expansion in basis |φi〉 are given by 〈φi|Ψ(0)〉.f) For the state found given the part (e) question, what is the probability at any time t of

measuring (i.e., forcing by the fundamental perturbation) the system into state

|+〉 =

(

10

)

?

HINT: Note a and b are in general complex.

g) Set a = 1 and b = 0 in the probability expression found in the part (f) answer. What isthe probability of measuring the system in state |+〉? in state |−〉? What is the systemdoing between the two states?

NOTE: The weird kind of oscillation between detectable states we have discussed is a simplemodel of neutrino oscillation. Just as an example, the detectable states could be the electronneutrino and muon neutrino and the particle oscillates between them. Really there are threeflavors of neutrinos and a three-way oscillation may occur. There is growing evidence thatneutrino oscillation does happen. (This note may be somewhat outdated due to that growth ofevidence.)

008 qfull 01700 2 5 0 moderate thinking: operator product ruleExtra keywords: Reference Ba-134

15. A function of an operator A can be defined by a power series

f(A, λ) =

∞∑

k=0

ak(λ)Ak ,

where λ is an example of c-number parameter of the function and convergence is guaranteed byfaith alone. For example, eλA means that

eλA =

∞∑

k=0

λk

k!Ak .

a) Show that a sensible definition of the derivative of f(A, λ) with respect to λ is

df

dλ=

∞∑

k=0

dak

dλAk .

HINT: Differentiate a general matrix element of f(A, λ).

b) Given the definition of the operator derivative of part (a), find product rule for operatorderivatives. HINT: Differentiate a general matrix element of f(A, λ)g(B, λ), where f andg are operator functions of general operators A and B.

008 qfull 01800 3 5 0 hard thinking: common eigensets, CSCOExtra keywords: See CDL-139–144

16. One can always construct a common basis (or common eigenset) for commuting observables(i.e., Hermitian operators that allow complete eigensets). Let us investigate this property.

a) First, an easy problem. Say you are given obserables A and B and they have a commoneigenset |ui

a,b〉 such that

A|uia,b〉 = a|ui

a,b〉 and B|uia,b〉 = b|ui

a,b〉 ,


where a and b are eigenvalues and i labels different states that are degenerate with respectto both eigenvalues: i.e., have the same a and b eigenvalues. Thus specifying a, b, and ifully specifies a state. Show that A and B must commute.

b) Now you are given [A,B] = 0 and the eigenset of A |uia〉, where i labels states that are

degenerate with respect to eigenvalues a. Show that you can construct a common eigensetfor A and B. HINT: Formally diagonalize B in the set |ui

a〉 making use of commutationto show that many matrix elements are zero.

c) Say A,B,C, . . . constitutes a complete set of commuting operators (i.e., a CSCO). Whatcan say about the common eigenset of A,B,C, . . .?

Chapt. 9 Time Evolution and the Heisenberg Representation


009 qmult 00100 1 4 1 moderate deducto-memory: constant of the motion1. What are the conditions for observable Q to be a constant of the motion.

a) [H,Q] = 0 and ∂Q/∂t = 0.b) [H,Q] 6= 0 and ∂Q/∂t 6= 0.c) [H,Q] > 0 and ∂Q/∂t > 0.d) [H,Q] < 0 and ∂Q/∂t < 0.e) [H,Q] ≥ 0 and ∂Q/∂t ≥ 0.


009 qfull 00100 3 5 0 tough thinking: time evolution, virial theoremExtra keywords: (Gr-117:3.53, Gr2005-126:3.31) No one remembers Dorothy Lamour.

1. Answer the following questions that lead up to the proof of the virial theorem. HINTS: Theanswers to the earlier parts help answering the later parts. But you can still answer some laterparts even if you don’t get all the earlier parts.

a) Given that e−iEnt/h−|φn〉, a STATIONARY STATE (i.e., an eigen-energy state) of atime-independent Hamiltonian with its time-dependence factor explicitly shown, show thatthe expectation value for this state of any time-independent operator A is a constant withrespect to time: i.e.,

d〈A〉dt

= 0 .

HINT: This is easy.

b) Given that |φn〉 is a stationary state of H and A is a general operator (i.e., it doesn’t haveto be an observable or Hermitian), show that

〈φn|[H,A]|φn〉 = 0 .

HINT: This is not so hard. Recall the formal definition of the Hermitian conjugate ofgeneral operator A is

〈α|A|β〉 = 〈β|A†|α〉∗ .

c) Prove that [A,BC] = [A,B]C +B[A,C] for general operators A, B, and C.

d) Prove [x, p] = ih−.

e) Prove that

[H,x] = − ih−mp .

68

Chapt. 9 Time Evolution and the Heisenberg Representation 69

f) Prove that

[H, p] = ih−(

∂V

∂x

)

.

g) Starting with the general time evolution equation (or general equation of motion) for generalobservable A

d〈A〉dt

=

⟨

∂A

∂t

⟩

+1

h−〈i[H,A]〉

show thatd〈xp〉dt

= 2〈T 〉 −⟨

x∂V

∂x

⟩

,

where T = p2/(2m) is the kinetic energy operator.

h) Show thatd〈xp〉dt

=d〈px〉dt

.

Whe quantity xp is the one-particle, one-dimensional virial operator. In classical physics,it would be a dynamical variable. HINT: This is easy.

i) Now for a STATIONARY STATE prove the 1-d virial theorem:

〈T 〉stationary =1

2

⟨

x∂V

∂x

⟩

stationary

.

HINT: Don’t forget part (a) and what the general equation of motion says.

j) Given potential V (x) ∝ xλ, show that the virial theorem reduces to

〈T 〉stationary =λ

2〈V 〉stationary .

009 qfull 01000 2 5 0 moderate thinking: Heis. Rep. evolutionExtra keywords: See Ba-137

2. Say A is an operator in the Schrodinger representation: this is the ordinary representation ofbeginning quantum mechanics. The Heisenberg representation of this operator for a systemwith Hamiltonian H is

A(t) = eiHt/h−Ae−iHt/h− ,

where exp(−iHt/h−) is operator function of H defined by

e−iHt/h− =∞∑

ℓ=0

1

ℓ!

(−iHth−

)ℓ

.

Show definitively thatdA(t)

dt=

1

ih−[A(t), H ] +

(

∂A

∂t

)

(t) ,

where the last term accounts for any implicit time dependence in A. HINTS: Take the timederivative of a general matrix element of A(t) with the general states expanded in the eigenketsof H . Note that in the Heisenberg representation states don’t have any time dependence: allthe time-dependence is in the operators.

009 qfull 01500 3 5 0 tough thinking: translation operatorExtra keywords: (Ba-145:2)

70 Chapt. 9 Time Evolution and the Heisenberg Representation

3. Here are some fun proofs to do that are all in the Schrodinger representation, except for thelast one.

a) Prove[~r, f(~p · ~u)] = ih−∇~pf(~p · ~u) ,

where ~u is just a c-number vector (i.e., a vector consisting of ordinary scalar constantcomponents) and the gradient operator is just a specially defined bit of formalism thatmeans differentiate with respect to components of ~p to form a gradient as if they wereordinary variables. HINT: Recall

f(~p · ~u) =

∞∑

ℓ=0

aℓ(~p · ~u)ℓ ,

where the aℓ are some scalar coefficients and convergence is assumed.

b) Using the part (a) result, prove that

ei~p·~u/h−~re−i~p·~u/h− = ~r + ~u .

c) Now prove that

|Ψ′〉 = e−i~p·~u/h−|Ψ〉

is the same state as |Ψ〉 translated by ~u. In the position representation

Ψ(~r )′ = e−i~p·~u/h−Ψ(~r ) = Ψ(~r − ~u) .

d) Given

|Ψ′〉 = e−i~p·~u/h−|Ψ〉 ,

where |Ψ′〉 and |Ψ〉 are in the Heisenberg representation, show that |Ψ′(t)〉 (the state |Ψ′〉in the Schrodinger representation) evolves according to

|Ψ′(t)〉 = e−i~p(−t)·~u/h−|Ψ(t)〉 ,

where ~p(−t) is the momentum operator in the Heisenberg representation at −t.

Chapt. 10 Measurement


010 qmult 00100 1 1 1 easy memory: fundamental perturbation1. In an ideal quantum mechanical measurement of an observable A:

a) the measurement always detects an EIGENVALUE of the observable and projects thesystem into an EIGENSTATE of the observable corresponding to that eigenvalue.

b) the measurement always detects an EXPECTATION VALUE of the observable andprojects the system into an EIGENSTATE of the observable.

c) the measurement always detects an EXPECTATION VALUE of the observable andprojects the system into an NON-EIGENSTATE of the observable.

d) the measurement always detects an 3 EIGENVALUES of the observable and projectsthe system into an NON-EIGENSTATE of the observable.

e) The measurement always detects an EXPECTATION VALUE of the observable andprojects the system into a STATIONARY STATE.


71

Chapt. 11 The Central Force Problem and Orbital Angular Momentum


011 qmult 00100 1 4 3 easy deducto-memory: central-force1. In a central-force problem, the magnitude of central force depends only on:

a) the angle of the particle.b) the vector ~r from the center to the particle.c) the radial distance r from the center to the particle.d) the magnetic quantum number of the particle.e) the uncertainty principle.

011 qmult 00200 1 1 2 easy memory: separation of variables2. The usual approach to getting the eigenfunctions of the Hamiltonian in multi-dimensions is:

a) non-separation of variables. b) separation of variables.c) separation of invariables. d) non-separation of invariables.e) non-separation of variables/invariables.

011 qmult 00210 1 1 3 easy memory: separation of variables3. Say you have a differential equation of two independent variables x and y and you want to look

for solutions that can be factorized thusly f(x, y) = g(x)h(y). Say then it is possible to reorderequation into the form

LHS(x) = RHS(y) ,

where LHS stands for left-hand side and RHS for right-hand side. Well LHS is explicitlyindependent of y and implicitly independent of x:

∂LHS

∂y= 0 and

∂LHS

∂x=∂RHS

∂x= 0 .

Thus, LHS is equal to a constant C and necessarily RHS is equal to the same constant C whichis called the constant of separation (e.g., Arf-383). The solutions for g(x) and h(y) can be foundseparately and are related to each other through C. The solutions for f(x, y) that cannot befactorized are not obtained, of course, by the described procedured. However, if one obtainscomplete sets of g(x) and h(y) solutions for the x-y region of interest, then any solution f(x, y)can be constructed at least to within some approximation (Arf-443). Thus, the generalizationof the described procedure is very general and powerful. It is called:

a) separation of the left- and right-hand sides. b) partitioning.c) separation of the variables. d) solution factorization. e) the King Lear method.

011 qmult 00300 1 4 2 easy deducto-memory: relative/cm reduction4. “Let’s play Jeopardy! For $100, the answer is: By writing the two-body Schrodinger equation

in relative/center-of-mass coordinates.”

How do you , Alex?

a) reduce a ONE-BODY problem to a TWO-BODY problemb) reduce a TWO-BODY problem to a ONE-BODY problem

72

Chapt. 11 The Central Force Problem and Orbital Angular Momentum 73

c) solve a one-dimensional infinite square well problemd) solve for the simple harmonic oscillator eigenvaluese) reduce a TWO-BODY problem to a TWO-BODY problem

011 qmult 00310 1 4 4 easy deducto-memory: reduced mass5. The formula for the reduced mass m for two-body system (with bodies labeled 1 and 2) is:

a) m = m1m2. b) m =1

m1m2. c) m =

m1 +m2

m1m2. d) m =

m1m2

m1 +m2.

e) m =1

m1.

011 qmult 00400 1 4 2 easy deducto memory: spherical harmonics 16. The eigensolutions of the angular part of the Hamiltonian for the central force problem are the:

a) linear harmonics. b) spherical harmonics. c) square harmonics.d) Pythagorean harmonics. e) Galilean harmonics.

011 qmult 00410 1 4 4 easy deducto-memory: spherical harmonics 2Extra keywords: mathematical physics

7. “Let’s play Jeopardy! For $100, the answer is: They form a basis or complete set for the 2-dimensional space of the surface a sphere which is usually described by the angular coordinatesof spherical polar coordinates.”

What are the , Alex?

a) Hermite polynomials b) Laguerre polynomialsc) associated Laguerre polynomials d) spherical harmonicse) Chebyshev polynomials

011 qmult 00420 1 4 3 easy deducto memory: spherical harmonic Y008. Just about the only spherical harmonic that people remember—and they really should remember

it too—is Y00 =:

a) eimφ. b) r2. c)1√4π

. d) θ2. e) 2a−3/2e−r/a.

011 qmult 00500 1 4 2 easy deducto-memory: spdf designations9. Conventionally, the spherical harmonic eigenstates for angular momentum quantum numbers

ℓ = 0, 1, 2, 3, 4, ...

are designated by:

a) a, b, c, d, e, etc.b) s, p, d, f , and then alphabetically following f : i.e., g, h, etc.c) x, y, z, xx, yy, zz, xxx, etc.d) A, C, B, D, E, etc.e) $@%&*!!

011 qmult 00510 1 4 3 easy deducto-memory: s electrons10. “Let’s play Jeopardy! For $100, the answer is: What the ℓ = 0 electrons (or zero orbital angular

momentum electrons) are called in spectroscopic notation.”

What are , Alex?

a) the Hermitian conjugates b) Herman’s Hermits c) s electrons d) p electronse) h electrons

74 Chapt. 11 The Central Force Problem and Orbital Angular Momentum


011 qfull 00100 2 5 0 moderate thinking: 2-body reduced to 1-body problemExtra keywords: (Gr-178:5.1)

1. The 2-body time-independent Schrodinger equation is

− h−2

2m1∇2

1ψ − h−2

2m2∇2

2ψ + V ψ = Etotalψ .

If the V depends only on ~r = ~r2 − r1 (the relative vector), then the problem can be separateinto two problems: a relative problem 1-body equivalent problem and a center-of-mass 1-bodyequivalent problem. The center of mass vector is

~R =m1~r1 +m2~r2

M,

where M = m1 +m2.

a) Determine the expressions for ~r1 and ~r2 in terms of ~R and ~r.

b) Determine the expressions for ∇21 and ∇2

2 in terms of ∇2cm (the center-of-mass Laplacian

operator) and ∇2 (the relative Laplacian operator). Then re-express kinetic operator

− h−2

2m1∇2

1 −h−2

2m2∇2

2

in terms of ∇2cm and ∇2. HINTS: The x, y, and z direction components of vectors can

all be treated separately and identically since x components of ~R and ~r) (i.e., X and x)depend only on x1 and x2, etc. You can introduce a reduced mass to make the transformedkinetic energy operator simpler.

c) Now separate the 2-body Schrodinger equation assuming V = V (~r ). What are the solutionsof the center-of-mass problem? How would you interpret the solutions of the relativeproblem? HINT: I’m only looking for a short answer to the interpretation question.

011 qfull 00200 2 3 0 moderate math: central-force azimuthal component solutionExtra keywords: solving the azimuthal component of the central force problem

2. In the central force problem, the separated azimuthal part of the Schrodinger equation is:

d2Φ

dφ2= −m2

ℓΦ ,

where −m2ℓ is the constant of separation for the azimuthal part. The constant has been

parameterized in terms of mℓ (which is not mass) since it turns out that for normalizable(and therefore physically allowed) solutions that m must be an integer. The mℓ quantity is thez-component angular momentum quantum number or magnetic quantum number (MEL-59;ER-240). The latter name arises since the z-components of the angular momentum manifestthemselves most noticeably in magnetic field phenomena.

a) Since the differential equation is second order, there should should be two independentsolutions for each value ofm2

ℓ . Solve for the general solution Φ for eachm2ℓ : i.e., the solution

that is a linear combination of the two independent solutions with undetermined coefficients.Note that writing the separation constant as m2

ℓ is so far just a parameterization andnothing yet demands that m2

ℓ be greater than zero or pure real. HINT: Use an exponential

Chapt. 11 The Central Force Problem and Orbital Angular Momentum 75

trial function with exponent ±(a+ ib) with a and b real. Also remember the special caseof m2

ℓ = 0.

b) The solutions are continuous and so that quantum mechanical requirement is met. Butanother one must be imposed for the azimuthal coordinate: the single-valuedness condition.Since we have no interpretation for multi-valuedness, we micropostulate that it doesn’thappen. Impose the single-valuedness condition on the generl solution

Φ = Ae(a+ib)φ +Be−(a+ib)φ ,

and show that a = 0 and mℓ must be an integer. Remember to consider the special casewhere mℓ = 0?

c) What are the eigenfunction solutions for the z-component of the angular momentumoperator

Lz =h−i

∂

∂φ.

What are the eigenvalues that satisfy single-valuedness and continuity? What is therelationship between these eigenfunction solutions and the azimuthal angle part of thehydrogenic atom wave functions?

d) Normalize the allowed eigensolutions of Lz Note these solutions are, in fact, conventionallyleft unnormalized: i.e., the coefficient of the special function that is the solution is left asjust 1. Normalization is conventionally imposed on the total orbital angular momentumsolutions, spherical harmonics.

011 qfull 01000 3 5 0 tough thinking: the nearly rigid rotator3. You have a 3-dimensional system consisting of two non-identical particles of masses m1 andm2. The two particles form a nearly rigid rotator. The relative time-independent Schrodingerequation for the system is:

[

− h−2

2µ

1

r2∂

∂r

(

r2∂

∂r

)

+L2

2µr2+ V (r)

]

Ψ(r, θ, φ) = EΨ(r, θ, φ) ,

where r, θ, and φ are the relative coordinates, µ = m1m2/(m1 +m2) is the reduced mass, andthe potential is

V (r) =

0 , for r ∈ [a− ∆a/2, a+ ∆a/2];∞ , otherwise.

a) Assume that ∆a is so much smaller than a that L2/(2µr2) ≈ L2/(2µa2). Now separatethe equation into radial and angular parts using Erad and Erot as the respective separationconstants: Erad +Erot = E. Let the radial solutions be R(r). You know what the angularsolutions should be. Write down the separated equations.

b) For the radial equation assume that r varies so much more slowly than R over the regionof non-infinite potential that

1

r2∂

∂r

(

r2∂R

∂r

)

≈ ∂2R

∂r2

in that region. Change the coordinate variable to x = r−(a−∆a/2) for simplicity: the non-infinite region of the potential then is then the x range [0,∆a]. With this approximationsolve for the radial eigenstates and eigen-energies. Normalize the eigenstates. HINTS:Holy deja vue all over again Batman, it’s the 1-dimensional infinite square well problem.Don’t mix up a and ∆a.

76 Chapt. 11 The Central Force Problem and Orbital Angular Momentum

c) Write down the eigenstates (just their general symbol, not expressions) and eigen-energyexpression for the rotational equation. What is the degeneracy of each eigen-energy?HINTS: You shouldn’t being trying to solve the equation. You should know what theeigenstates are.

d) Write the general expression for the total wave function. How many quantum numbersdoes it depend on?

e) Write down the general expression for the total energy. Which causes a greater change inenergy: a change of 1 in the quantum number controling the radial energy or a changeof 1 in the quantum number controling the rotational energy? Remember ∆a << a byassumption.

f) Sketch the energy level diagram.

Chapt. 12 The Hydrogenic Atom


012 qmult 00050 1 1 1 easy memory: hydrogen atom, 2-body1. The hydrogen atom is the simplest of all neutral atoms because:

a) it is a 2-body system. b) it is a 3-body system. c) it has no electrons.d) it has many electrons. e) hydrogen is the most abundant element in the universe.

012 qmult 00100 1 1 3 easy memory: radial wave function requirements2. What basic requirements must the radial part of hydrogenic atom wave function meet in order

to be a physical radial wave function?

a) Satisfy the radial part of the Schrodinger equation and grow exponentially as r → ∞.b) Not satisfy the radial part of the Schrodinger equation and grow exponentially as r → ∞.c) Satisfy the radial part of the Schrodinger equation and be normalizable.d) Not satisfy the radial part of the Schrodinger equation and be normalizable.e) None at all.

012 qmult 00190 1 1 2 easy memory: hydrogen wave functions3. The hydrogenic atom eigenstate wave functions contain a factor that causes them to:

a) increase exponentially with radius. b) decrease exponentially with radius.c) increase logarithmically with radius. d) increase quadratically with radius.e) increase linearly with wavelength.

012 qmult 00200 1 4 1 easy deducto-memory: associated Laguerre polyn.4. What special functions are factors in the radial part of the of the hydrogenic atom eigenstate

wave functions?

a) The associated Laguerre polynomials. b) The unassociated Laguerre polynomials.c) The associated Jaguar polynomials. d) The unassociated jaguar polynomials.e) The Hermite polynomials.

012 qmult 01000 1 4 1 easy deducto-memory: atomic spectroscopy5. Almost all would agree that the most important empirical means for learning about atomic

energy eigenstates is:

a) spectroscopy. b) microscopy. c) telescopy. d) pathology. e) astrology.


012 qfull 00100 1 1 0 easy memory: separation of two body problem.1. The full Schrodinger equation for the hydrogenic atom is a function of two positions, one for

the electron and one for the nucleus. What must one do to turn the problem into a central forceproblem for one body?

77

78 Chapt. 12 The Hydrogenic Atom

012 qfull 00200 2 5 0 moderate thinking: how does ¡r¿ vary with n?2. How does the mean radius (expectation value radius) 〈r〉nℓm for the hydrogenic atom vary with

increasing n (i.e., with increasing energy)?

012 qfull 00300 2 1 0 moderate memory: H atom quantum numbers3. What are the 3 quantum numbers of the hydrogenic atom derived from the spatial Schrodinger

equation?

012 qfull 00400 2 1 0 moderate memory: s electron polar plot4. Sketch the polar plot for an s electron (i.e., an ℓ = 0 electron)?

012 qfull 00500 2 5 0 moderate thinking: rotating or standing wave functions5. Are the hydrogenic wave functions Ψnlm rotating wave or standing wave functions?

012 qfull 00600 2 5 0 moderate thinking: rotating or standing wave functions6. Can there be hydrogenic atom stationary-state standing wave functions?

012 qfull 00700 2 5 0 moderate thinking: what is the Bohr magneton?7. What is the Bohr magneton?

012 qfull 00800 2 5 0 moderate thinking: atomic magnetic moments8. Why should an atom have a magnetic moment?

012 qfull 00900 1 3 0 easy math: first 4 Laguerre polynomials keyword first 4 Laguerre polynomials,Rodrigues’s formula

9. Using Rodrigues’s formula for Laguerre polynomials (NOT Legendre polynomials) determinethe first 4 Laguerre polynomials.

012 qfull 01000 3 3 0 tough thinking : separation of external potentialExtra keywords: separation of external potential, 1st order expansion

10. Consider the initial hydrogenic-atom Schrodinger equation where the position variables are stillfor the nucleus and electron. Say we add perturbation potentials Vn(~r n) for the nucleus andVe(~r e) for the electron. We further specify that these perturbation potentials vary only linearlywith position. How would one have to treat these potentials in order to transform to the center-of-mass/relative coordinate system and separate the Schrodinger equation? HINTS: Have youheard of the Taylor’s series? You’ll have to express the ~rn and ~r e in terms of relative and centerof mass coordinates.

012 qfull 01200 2 3 0 moderate math: s electron in nucleusExtra keywords: (Gr-142:4.14)

11. Let us consider the probability that the electron of a hydrogenic atom in the ground state willbe in the nucleus. Recall the wave function for ground state is

Ψ100(~r ) = R10(r)Y00(θ, φ) = 2a−3/2e−r/a × 1√4π

(Gr2005-154), where a = aBohr[me/(mZN)]: aBohr ≈ 0.529A is the Bohr radius, ZN is thenuclear charge, me is the electron mass, and m is the reduced mass of the actual hydrogenicatom.

a) First assume that the wave function is accurate down to r = 0. It actually can’t be ofcourse. The wave function was derived assuming a point nucleus and the nucleus is, infact, extended. However, the extension of the nucleus is of order 105 times smaller thanthe Bohr radius, and so the effect of a finite nucleus is a small perturbation. Given that

Chapt. 12 The Hydrogenic Atom 79

the nuclear radius is b, calculate the probability of finding the electron in the nucleus. Useǫ = b/(a/2) = 2b/a to simplify the formula. HINT: The formula

g(n, x) =

∫ x

0

e−ttn dt = n!

(

1 − e−xn∑

ℓ=0

xℓ

ℓ!

)

could be of use.

b) Expand the part (a) answer in ǫ power series and show to lowest non-zero order that

P (r < b, ǫ << 1) =1

6ǫ3 =

4

3

(

b

a

)3

.

c) An alternate approach to find the probability of the electron being in the nucleus is assumeΨ(~r ) can be approximated by Ψ(0) over nucleus. Thus

P (r < b) ≈(

4π

3

)

b3|Ψ(0)|2 .

Is this result consistent with the part (b) answer?

d) Assume b ≈ 10−15 m and a = 0.5 × 10−10 m. What is the approximate numerical valuefor finding the electron in the nucleus? You can’t interpret this result as “the fraction ofthe time the electron spends in the nucleus”. Nothing in quantum mechanics tells us thatthe electron spends time definitely anywhere. One should simply stop with what quantummechanics gives: the result is the probability of finding the electron in nucleus.

012 qfull 01300 3 5 0 tough thinking: derivation of quantum J currentExtra keywords: derivation of quantum ~J current, correspondence principle

12. Let’s see if we can derive the probability current density from the correspondence principle.Note that the classical current density is given by ~jcl = ~vclρcl. (a) First off we have to figureout what the quantum mechanical ρ and ~j are classified as in quantum mechanics? Are theyoperators or wave functions or expectation values or are they just their own things? Wellthey may indeed be just their own things, but one can interpret them as belonging to oneof the three mentioned categories. Which? (b) Well now that part (a) is done we can usethe correspondence principle to find an operator corresponding to classical ~jcl. What are thethe appropriate operators to replace the classical ρcl and ~vcl with (i.e., how are ρcl and ~vclquantized)? (c) Have you remembered the quantization symmetrization rule? (d) Now go to itand derive the quantum mechanical ~j. You might find the 3-d integration-by-parts rule handy:

∫

V

Ψ∇χdV =

∫

A

Ψχd ~A−∫

V

∇ΨχdV ,

where∫

V is for integral over all volume V and∫

A is for integral over all vectorized surface areaof volume V .

012 qfull 02100 1 3 0 easy math: positronium solution13. Positronium is an exotic atom consisting of an electron and its antiparticle the positron. It was

predicted to exist in 1934 (or even earlier) shortly after the positron was discovered in 1932.Positronium was experimentally discovered in 1951. Positronium cannot exist long because theelectron and positron will mutually annihilate usually producing two γ-rays although more γ-rays are possible since there are multiple modes of annihilation. Positronium frequenctly formsin exited states and decays by radiative transitions to the ground state unless it annihilatesfirst by some mode. Positronium transition spectra and annihilation γ-ray spectra provide a

80 Chapt. 12 The Hydrogenic Atom

fine test of quantum mechanics and quantum electrodynamics. Neglecting annihilation effects,spin effects, and relativistic effects, positronium to first order is Schrodinger-solution hydrogenicatom. We just consider this simplified positronium in this problem.

a) What are the positronium total mass and reduced mass?

b) What is the formula for the energy of the energy levels of positronium?

c) How does the emitted/absorbed photon of a positronium line transition (i.e., transitionbetween energy levels) compare to the corresponding line transition photon of theSchrodinger-solution HYDROGEN atom? By “corresponding”, we mean that thephotons result from transitions that have the same initial and final principal quantumnumbers.

Chapt. 13 General Theory of Angular Momentum


013 qmult 00100 1 1 4 easy memory: ang. mom. commutation relations1. The fundamental angular momentum commutation relation and a key corollary are, respectively:

a) [Ji, Jj ] = 0 and [J2, Ji] = Ji. b) [Ji, Jj ] = Jk and [J2, Ji] = 0.c) [Ji, Jj ] = 0 and [J2, Ji] = 0. d) [Ji, Jj ] = ih−εijkJk and [J2, Ji] = 0.e) [xi, pj ] = ih−δij , [xi, xj ] = 0, and [pi, pj ] = 0.

013 qmult 00910 1 1 3 easy memory: vector model2. In the vector model for angular momentum of a quantum system with the standard axis for the

eigenstates being the z axis, the particles in the eigenstates are thought of as having definitez-components of angular momentum mj h− and definite total angular momenta of magnitude√

j(j + 1)h−, where j can stand for orbital, spin, or total angular momentum quantum numberand mj is the z-component quantum number. Recall j can be only be integer or half-integerand there are 2j+ 1 possible values of mj given by −j,−j+ 1, . . . , j − 1, j. The x-y component

of the angular momementum has magnitude√

j(j + 1) −m2j h−, but it has no definite direction.

Rather this component can be thought of as pointing all x-y directions in simultaneous: i.e., itis in a superposition state of all direction states. Diagramatically, the momentum vectors canbe represented by

a) cones with axis aligned with the x-axis. b) cones with axis aligned with the y-axis.c) cones with axis aligned with the z-axis.d) cones with axis aligned with the x-y-axis. e) the cones of silence.

013 qmult 01000 1 1 5 easy memory: rigid rotator eigen-energies3. For a rigid rotator the rotational eigen-energies are proportional to:

a) ℓh−. b) ℓ2h−2. c) h−2

/[ℓ(ℓ+ 1)]. d) h−2/ℓ2. e) ℓ(ℓ+ 1)h−2

.

013 qmult 02000 1 1 1 easy memory: added ang. mom. operators4. Does the fundamental commutation relation for angular momentum operators (i.e., [Ji, Jj ] =ih−εijkJk) apply to angular momentum operators formed by summation from angular momentumoperators applying to individual particles or to spatial and spin degrees of freedom? The answeris:

a) Yes. b) No. c) Maybe. d) All of the above. e) None of the above.

013 qmult 02100 1 4 5 easy deducto-memory: Clebsch-Gordan coefficients5. “Let’s play Jeopardy! For $100, the answer is: The name for the coefficients used in the

expansion of a total angular momentum state for 2 angular momentum degrees of freedom interms of products of individual angular momemtum states.”

What are the , Alex?

a) Racah W coefficients b) Wigner 6j symbols c) Buck-Rogers coefficientsd) Flash-Gordon coefficients e) Clebsch-Gordan coefficients

81

82 Chapt. 13 General Theory of Angular Momentum

013 qmult 02200 1 4 5 easy deducto-memory: Clebsch-Gordan m rule6. “Let’s play Jeopardy! For $100, the answer is: In constructing a set of |j1j2jm〉 states from

a set of |j1j2m1m2〉 states using Clebsch-Gordan coefficients, this is a strict constraint on thenon-zero coefficients.”

What is the rule , Alex?

a) of complete overtures b) of incomplete overtures c) m = m21 +m2

2

d) m = m1 −m2 e) m = m1 +m2


013 qfull 00090 2 5 0 moderate math: kroneckar delta, Levi-Civita1. There are two symbols that are very useful in dealing with quantum mechanical angular

momentum and in many other contexts in physics: the Kronecker delta:

δij =

1 , i = j;0 , i 6= j;

and the Levi-Civita symbol

εijk =

1 , if ijk is a cyclic permutation of 123 (3 cases);−1 , if ijk is an anticyclic permutation of 123 (3 cases);0 , if any two indices are the same.

NOTE: Leopold Kronecker (1823–1891) was a German mathematician although born in whatis now Poland. Tullio Levi-Civita (1873–1941) was an Italian mathematician: the “C” in Civitais pronounced “ch”.

a) Prove δijδik = δjk, where we are using Einstein summation here and below, of course.

b) Now the toughie. Proveεijkεiℓm = δjℓδkm − δjmδkℓ .

HINTS: I know of no simple one or two line proof. The best I’ve ever thought of wasto consider cases where jkmℓ span 3, 1, and 2 distinct values and to show that the twoexpressions are equal in all cases.

c) Now the cinchy one. Proveεijkεijm = 2δkm .

d) What does εijkεijk equal? Note there is Einstein summation on all indices now.

013 qfull 00100 2 5 0 moderate thinking: angular momentum operator identities2. Prove the following angular momentum operator identities. HINT: Recall the fundamental

angular momentum commutator identity,

[Ji, Jj ] = ih−εijkJk , and the definition J± ≡ Jx ± iJy .

a) [Ji, J2] = 0.

b) [J2, J±] = 0.c) [Jz, J±] = ±h−J±.

d) J†±J± = J∓J± = J2 − Jz(Jz ± h−).

Chapt. 13 General Theory of Angular Momentum 83

e)

Jx =1

2(J+ + J−) and Jy =

1

2i(J+ − J−) .

f) [J+, J−] = 2h−Jz.g)

J2

x

y = ±1

4

(

J2+ + J2

− ± J+, J−)

,

where the upper case is for J2x and the lower case for J2

y and where recall that A,B =AB +BA is the anticommutator of A and B.

h)

J2 =1

2J+, J− + J2

z .

013 qfull 00200 2 3 0 mod math: diagonalization of Jx for 3-dExtra keywords: diagonalization of the J-x angular momentum matrix for 3-d

3. The x-component angular momentum operator matrix in a three-dimensional angularmomentum space expressed in terms of the z-component orthonormal basis (i.e., the standardbasis with eigenvectors |1〉, |0〉, and | − 1〉) is:

Jx =h−√2

0 1 01 0 10 1 0

(Co-659) and yes the 1/√

2 factor is correct. Is this matrix Hermitian? Diagonalize this matrix:i.e., solve for its eigenvalues and normalized eigenvectors (written in terms of the standard basisket eigenvectors) or, if you prefer in column vector form. Note the solution is somewhat simplerif you solve the reduced eigen problem. Just divide both sides of the eigen equation by h−/

√2

and solve for the reduced eigenvalues. The physical eigenvalues are the reduced ones timesh−/

√2. Verify that the eigenvectors are orthonormal.NOTE: Albeit some consider it a sloppy notation since kets and bras are abstract vectors

and columns vectors are from a concrete representation, its concretely useful to equate them attimes. In the present case, the kets equate like so

|1〉 =

100

, |0〉 =

010

, | − 1〉 =

001

,

and the bras, like so

〈1| = (1, 0, 0)∗ , 〈0| = (0, 1, 0)∗ , 〈−1| = (0, 0, 1)∗ .

013 qfull 00300 2 3 0 mod math: diagonalization of J-y for 3-dExtra keywords: diagonalization of the J-y angular momentum matrix for 3-d

4. The y-component angular momentum operator matrix in a three-dimensional angularmomentum space expressed in terms of the z-component orthonormal basis (i.e., the standardbasis with eigenvectors |1〉, |0〉, and | − 1〉) is:

Jy =h−√2

0 −i 0i 0 −i0 i 0


(Co-659) and yes the 1/√

2 factor is correct. Is this matrix Hermitian? Diagonalize this matrix:i.e., solve for its eigenvalues and normalized eigenvectors (written in terms of the standard basiskets) or, if you prefer in column vector form. Verify that the eigenvectors are orthonormal.

Note the solution is somewhat simpler if you solve the reduced eigen problem. Just divideboth sides of the eigen equation by h−/

√2 and solve for the reduced eigenvalues. The physical

eigenvalues are the reduced ones times h−/√

2.

NOTE: Albeit some consider it a sloppy notation since kets and bras are abstract vectorsand columns vectors are from a concrete representation, its concretely useful to equate them attimes. In the present case, the kets equate like so

|1〉 =

100

, |0〉 =

010

, | − 1〉 =

001

,

and the bras, like so

〈1| = (1, 0, 0)∗ , 〈0| = (0, 1, 0)∗ , 〈−1| = (0, 0, 1)∗ .

013 qfull 00400 2 3 0 mod math: angular momemtum eqn. of motionExtra keywords: (Gr-150:4.21) torque

5. Let’s consider the angular momentum equation of motion in in the context of quantummechanics.

a) Prove that

d〈~L 〉dt

= 〈~τ 〉 ,

where ~L = ~r × ~p is the angular momentum operator and ~τ = ~r × (−∇V ) is the torqueoperator.

b) Then prove that

d〈~L 〉dt

= 0

for any central potential system: i.e., a system where the potential depends on radius alone.

HINTS: You’ll need to use the general time evolution equation—or equation of motion orderivative of expectation value: whatever one calls it—people do seem to avoid giving it a name.Then you will need to work out a commutation relation with a cross product operator. Thereare two approaches. First, show what the commutation relation is component by component.But that’s for pedestrians. The second way is to use the Levi-Civita symbol with the Einsteinsummation rule to prove the all commutation relations simultaneously. Part (a) is most easilydone using Cartesian coordinates and part (b) using spherical polar coordinates.

013 qfull 00500 2 3 0 moderate thinking: orbital angular momentumExtra keywords: expectation values, standard deviations, quantum and classical analogs

6. Consider a spinless particle in an eigenstate |ℓ,m〉 of the L2 and Lz operators: ℓ is the L2

quantum number and m the Lz quantum number. The set of |ℓ,m〉 states are a completeorthonormal set for angular coordinates. Recall

L2|ℓ,m〉 = ℓ(ℓ+ 1)h−2|ℓ,m〉 ,Lz|ℓ,m〉 = mh−|ℓ,m〉 ,L±|ℓ,m〉 = h−

√

ℓ(ℓ+ 1) −m(m± 1)|ℓ,m± 1〉 ,and


L± = Lx ± iLy .

a) Solve for expectation values 〈Lx〉, and 〈Ly〉, and standard deviations ∆Lx and ∆Ly.HINTS: You will need expressions for Lx and Ly in terms of the given operators. Alsothe everything can be done by operator algebra: there is no need to bring in the sphericalharmonics or particular representations of the operators.

b) Let us now see if there are classical analogs to the results in part (a). Let classical

Lz = mh− ,

Lx = h−√

ℓ(ℓ+ 1) −m2 cos(φ)

and

Ly = h−√

ℓ(ℓ+ 1) −m2 sin(φ) ,

where φ is the azimuthal angle of the angular momentum vector. Note L2x + L2

y + L2z =

ℓ(ℓ+1). Now solve for the classical 〈Lx〉 and 〈Ly〉, and the classical ∆Lx and ∆Ly assuming(i) that φ has a random uniform distribution the range [0, 2π] and (ii) that φ = wt whereω is a constant angular frequency.

013 qfull 00600 2 5 0 moderate thinking: orb. ang. mon. commutator proofs7. You are given the basic commutator identity [ri, pj ] = ih−δij and the correspondence principle

result ~L = ~r × ~p.

a) Prove [Li, rj ] = ih−εijkrk.

b) Prove [Li, pj] = ih−εijkpk.

c) Prove [Li, Lj] = ih−εijkLk. HINT: Remember the old subtract and add the same thingtrick.

d) Prove [Li, qjqj ] = 0, where ~q is any of ~r, ~p, and ~L.

e) Prove [Li, Qj ] = ih−εijkQk with ~Q = A~qB, where A and B are any scalar combination of

~r, ~p, and ~L: e.g., A = r2kp4mL2n . . ., where k, m, and n are integers.

f) Prove [α · ~L, ~Q] = ih−~Q× α, where α is a constant c-number unit vector.

g) Show that

d ~Q

dα=

i

h−[α · ~L, ~Q]

is the differential equation for a right-hand-rule rotation by α of operator ~Q about the axisin the direction α. HINT: I’m not looking for mathematical rigor—but if you can do thatit’s OK.

h) Show that the solution of

d ~Q

dα=

i

h−[α · ~L, ~Q]

is~Q = ei~L·~α/h− ~Q0e

−i~L·~α/h− ,

where ~α = αα is a general angle in vector form and ~Q0 is the inital operator ~Q.

013 qfull 02000 1 5 0 easy thinking: ang. mom. fundamental commutationExtra keywords: for addition. Reference Ba-332.


8. The fundamental commutation relation of angular momentum can be generalized for multipledegrees of freedom. The degrees of freedom could be the angular momenta of multiple particlesor the spatial and spin angular momenta of a particle or combinations thereof. Say we havedegrees of freedom f and g, then the relation is

[Jfi, Jgj ] = δfgih−εijkJfk .

We see that component operators refering to different degrees of freedom commute: this is trueeven in the case of indistinguishable particles. The total angular momentum operator ~J for a setof degrees of freedom with individual angular momentum operators ~Jf is, by the correspondenceprinciple,

~J =∑

f

~Jf .

a) Prove that the fundamental commutation relation holds for the components of ~J : i.e., prove

[Ji, Jj ] = ih−εijkJk .

What does this result imply for summed angular momenta?

b) Now let ~J = ~J1 + ~J2. ProveJ± = J1± + J2± .

c) ProveJ2 = J2

1 + J22 + J1+J2− + J1−J2+ + 2J1zJ2z .

013 qfull 02100 3 5 0 hard thinking: Clebsch-Gordan ell plus 1/2Extra keywords: See Ba-341 and CDL-1020

9. One special case of great interest for which general formulae can be found for all Clebsch-Gordancoefficients is that of a general angular momentum added to a spin 1/2 angular momentum. Letthe original angular momentum operators be labeled J2

1 , J1z, J22 , and J2z : the corresponding

eigenvalues are j1(j1 + 1), m1, (1/2)(1/2 + 1), and ±1/2. The set of product states of theoriginal operators is |j1, 1/2,m1,m2 = ±1/2〉. The summed operators are J2 and Jz:

the corresponding eigenvalues are j and m. The set of eigenstates of ~J21 , J2

2 , J2, and Jz is|j1, 1/2, j = j1 ± 1/2,m〉. The expression for the Clebsch-Gordan coefficients is

〈j1, 1/2,m1,m2 = ±1/2|j1, 1/2, j = j1 ± 1/2,m〉 .

a) For a given j1 what are the possible j values?

b) Consider the trivial subspace for j1 = 0. What are all the Clebsch-Gordan coefficients forthis subspace.

c) Now consider the general subspace for j1 ≥ 1/2. First find the expression for the summedstate with the largest m value. HINT: Recall

〈j1, j2,m1,m2|Jz|j1, j2, j,m〉 = 〈j1, j2,m1,m2|(J1z + J2z)|j1, j2, j,m〉 ,and so

m〈j1, j2,m1,m2|j1, j2, j,m〉 = (m1 +m2)〈j1, j2,m1,m2|j1, j2, j,m〉 .

Thus, the Clebsch-Gordan coefficient is zero unless m = m1 +m2.

d) Determine the expression for

|j1, 1/2, j = j1 + 1/2,m = j1 − 1/2〉 .


e) Now show that general expression for Clebsch-Gordan coefficient

〈j1, 1/2,m1 = m− 1/2,m2 = 1/2|j1, 1/2, j = j1 + 1/2,m〉

is given by

〈j1, 1/2,m1 = m− 1/2,m2 = 1/2|j1, 1/2, j = j1 + 1/2,m〉

=

√

j1(j1 + 1) − (m+ 1/2)(m− 1/2)

(j1 + 1/2)(j1 + 3/2)− (m+ 1)m

√

j1(j1 + 1) − (m+ 3/2)(m+ 1/2)

(j1 + 1/2)(j1 + 3/2) − (m+ 2)(m+ 1). . .

√

j1(j1 + 1) − j1(j1 − 1)

(j1 + 1/2)(j1 + 3/2)− (j1 + 1/2)(j1 − 1/2).

HINT: What is mostly needed is a word argument.

f) Now show that

√

j1(j1 + 1) − (m+ 1/2)(m− 1/2)

(j1 + 1/2)(j1 + 3/2) − (m+ 1)m

√

j1(j1 + 1) − (m+ 3/2)(m+ 1/2)

(j1 + 1/2)(j1 + 3/2)− (m+ 2)(m+ 1). . .

√

j1(j1 + 1) − j1(j1 − 1)

(j1 + 1/2)(j1 + 3/2) − (j1 + 1/2)(j1 − 1/2)

=

√

j1 +m+ 1/2

2j1 + 1.

HINTS: Simplify√

j1(j1 + 1) − (m+ 1/2)(m− 1/2)

(j1 + 1/2)(j1 + 3/2) − (m+ 1)m

by dividing top and bottom by a common factor. You might try (Aj1 +Bm+ . . .)(Cj1 +Dm+ . . .) factorizations of the top and bottom.

g) Now show that the general expressions for the Clebsch-Gordan coefficients are

〈j1, 1/2,m1 = m∓ 1/2,m2 = ±1/2|j1, 1/2, j = j1 + 1/2,m〉 =

√

j1 ±m+ 1/2

2j1 + 1

〈j1, 1/2,m1 = m∓ 1/2,m2 = ±1/2|j1, 1/2, j = j1 − 1/2,m〉 = ∓√

j1 ∓m+ 1/2

2j1 + 1.

HINTS: Use the parts (e) and (f) answers and the normalization and orthogonalityconditions.

h) The operator ~J1 · ~S turns up in the spin-orbit interaction where ~J1 = ~L. Show that the

summed states |j1, 1/2, j = j1±1/2,m〉 are eigenstates of ~J1 · ~S. What are the eigenvalues?

Chapt. 14 Spin


014 qmult 00100 1 4 5 easy deducto-memory: spin definedExtra keywords: mathematical physics

1. “Let’s play Jeopardy! For $100, the answer is: It is the intrinsic angular momentum of afundamental (or fundamental-for-most-purposes) particle. It is invariant and its quantumnumber s is always an integer or half-integer.

What is , Alex?

a) rotation b) quantum number c) magnetic momentd) orbital angular momentum e) spin

014 qmult 00110 1 4 1 easy deducto-memory: Goudsmit and Uhlenbeck, spinExtra keywords: Don’t abbreviate: it ruins the joke

2. “Let’s play Jeopardy! For $100, the answer is: Goudsmit and Ulhenbeck.”

a) Who are the original proposers of electron spin in 1925, Alex?b) Who performed the Stern-Gerlach experiment, Alex?c) Who are Wolfgang Pauli’s evil triplet brothers, Alex?d) What are two delightful Dutch cheeses, Alex?e) What were Rosencrantz and Gildenstern’s first names, Alex?

014 qmult 00120 1 1 1 easy memory: spin magnitude3. A spin s particle’s angular momentum vector magnitude (in the vector model picture) is

a)√

s(s+ 1)h−. b) sh− c)√

s(s− 1)h− d) −sh− e) s(s+ 1)h−2

014 qmult 00130 1 1 5 easy memory: eigenvalues of spin 1/2 particle4. The eigenvalues of a COMPONENT of the spin of a spin 1/2 particle are always:

a) ±h−. b) ± h−3

. c) ± h−4

. d) ± h−5

. e) ± h−2

.

014 qmult 00130 1 1 2 easy memory: eigenvalues of spin s particle5. The quantum numbers for the component of the spin of a spin s particle are always:

a) ±1. b) s, s− 1, s− 2, . . . ,−s+ 1,−s. c) ±1

2. d) ±2. e) ±1

4.

014 qmult 00140 1 4 2 easy deducto-memory: spin and environment6. Is the spin (not spin component) of an electron dependent on the electron’s environment?

a) Always.b) No. Spin is an intrinsic, unchanging property of a particle.c) In atomic systems, no, but when free, yes.d) Both yes and no.e) It depends on a recount in Palm Beach.

88

Chapt. 14 Spin 89

014 qmult 00400 1 4 5 easy deducto-memory: spin commutation relation7. “Let’s play Jeopardy! For $100, the answer is:

[Si, Sj] = ih−εijkSk .”

What is , Alex?

a) the spin anticommutator relation b) an implicit equation for εijk

c) an impostulate d) an inobservablee) the fundamental spin commutation relation

014 qmult 00500 1 4 2 easy deducto-memory: Pauli spin matrices8. “Let’s play Jeopardy! For $100, the answer is:

σx =

(

0 11 0

)

, σy =

(

0 −ii 0

)

, σz =

(

1 00 −1

)

.”

What are , Alex?

a) dimensioned spin 1/2 matrices b) the Pauli spin matricesc) the Pauli principle matrices d) non-Hermitian matricese) matrix-look-alikes, not matrices

014 qmult 00600 1 1 1 easy memory: spin anticommutator relation9. The expression

σi, σj = 2δij1

is

a) the Pauli spin matrix anticommutator relation.b) the Pauli spin matrix commutator relation.c) the fundamental spin commutator relation.d) the covariance of two standard deviations. e) an oddish relation.

014 qmult 00700 1 1 2 easy memory: spin rotation DE10. The expression

d(~S · n)

dα= − i

h−[~S · α, ~S · n]

is a differential equation for

a) α.

b) the spin operator ~S · n as a function of rotation angle ~α.c) n.

d) the translation of the spin operator ~S.e) none of the above.

014 qmult 00800 1 4 5 easy deducto-memory: spin rotation operator11. “Let’s play Jeopardy! For $100, the answer is:

U(α) = e−i~S·~α/h− = e−i~σ·~α/2 .”

a) What the Hermitian conjugate of U(−2α), Alex?b) What is a bra, Alex?c) What is a spin 1/2 eigenstate, Alex?d) What is the NON-UNITARY operator for the right-hand rule rotation of a spin 1/2

state by an angle α about the axis in the direction α, Alex?

90 Chapt. 14 Spin

e) What is the UNITARY operator for the right-hand rule rotation of a spin 1/2 state byan angle α about the axis in the direction α, Alex?

014 qmult 00900 1 1 3 easy memory: space and spin operators commute12. A spatial operator and a spin operator commute:

a) never. b) sometimes. c) always. d) always and never. e) to the office.

014 qmult 01000 2 1 4 moderate memory: joint spatial-spin rotation13. The operator

U(α) = e−i ~J·~α/h−

a) creates a spin 1/2 particle.b) annihilates a spin 1/2 particlec) left-hand-rule rotates both the space and spin parts of states by an angle α about an axis

in the α direction.d) right-hand-rule rotates both the space and spin parts of states by an angle α about an axis

in the α direction.e) turns a state into a U-turn state.

014 qmult 01100 1 4 5 easy deducto-memory: spin-magnetic interaction14. “Let’s play Jeopardy! For $100, the answer is:

~µ = gq

2m~J , ~F = ∇(~µ · ~B) , ~τ = ~µ× ~B , H = −~µ · ~B .”

a) What are Maxwell’s equations, Alex?b) What are incorrect formulae, Alex?c) What are classical formulae sans any quantum mechanical analogs, Alex?d) What are quantum mechanical formulae sans any classical analogs, Alex?e) What are formulae needed to treat the interaction of angular momentum of a particle and

magnetic field in classical and quantum mechanics, Alex?

014 qmult 01200 1 1 2 easy memory: Bohr magneton15. What is

µB =eh−2me

= 9.27400915(23)× 10−24 J/T = 5.7883817555(79)× 10−5 eV/T ?

a) The nuclear magneton, the characteristic magnetic moment of nuclear systems.b) The Bohr magneton, the characteristic magnetic moment of electronic systems.c) The intrinsic magnetic dipole moment of an electron.d) The coefficient of sliding friction.e) The zero-point energy of an electron.

014 qmult 01210 1 1 3 easy memory: g factor g-factor16. The g factor in quantum mechanics is the dimensionless factor for some system that multiplied

by the appropriate magneton (e.g., Bohr magneton for electron systems) times the angularmomentum of the system divided by h− gives the magnetic moment of the system. Sometimesthe sign of the magnetic moment is included in the g factor and sometimes it is just shownexplicitly. The modern way seems to be to include it, but yours truly finds that awkward andso for now yours truly doesn’t do it. For the electron, the intrinsic magnetic moment operatorassociated with intrinsic spin is given by

~µop = −gµB

~Sop

h−,

Chapt. 14 Spin 91

where µB is the Bohr magneton and Sop is the spin vector operator. What is g for the intrinsicmagnetic moment operator of an electron to modern accuracy?

a) 1. b) 2. c) 2.0023193043622(15). d) 1/137. e) 137.

014 qmult 01210 1 1 4 easy memory: magnetic moment precessionExtra keywords: The precession is also called Larmor precession (En-114)

17. An object in a uniform magnetic field with magnetic moment due to the object’s angularmomentum will both classically and quantum mechanically:

a) Lancy progress. b) Lorenzo regress. c) London recess. d) Larmor precess.e) Lamermoor transgress.

014 qmult 01300 1 1 3 easy memory: Zeeman effectExtra keywords: See Ba-312 and Ba-466–468

18. What is an effect that lifts the angular momentum component energy degeneracy of atoms?

a) The spin-orbit effect.b) The Paschen-Back effect or, for strong fields, the Zeman effect.c) The Zeeman effect or, for strong fields, the Paschen-Back effect.d) The Zimmermann effect.e) The Zimmermann telegram.

014 qmult 01500 1 4 1 easy deducto-memory: spin resonanceExtra keywords: See Ba-317

19. “Let’s play Jeopardy! For $100, the answer is: the effect in which a weak sinusoidal radiofrequency magnetic field causes a particle with spin to precess about a direction perpendicularto strong uniform magnetic field that separates the spin component energy levels of the particlein energy.”

a) What is spin magnetic resonance, Alex?b) What is spin magnetic presence, Alex?c) What is the preferred spin effect, Alex?d) What is the dishonored spin effect, Alex?e) What is the Zeeman effect, Alex?

014 qmult 01600 1 1 3 easy memory: spin resonance field20. In spin magnetic resonance you can replace a rotating magnetic field by a sinusoidal one if you

can neglect or filter:

a) magnetic fields altogether.b) precession altogether.c) the very HIGH frequency effects of the sinusoidal field.d) spin altogether.e) the very LOW frequency effects of the sinusoidal field.

014 qmult 01700 1 4 5 easy deducto-memory: atomic clock21. “Let’s play Jeopardy! For $100, the answer is: The simplest of these consists of a beam of

spinned particles that passes through two cavities each with crossed constant and sinusoidalmagnetic fields.”


a) Stern-Gerlach experiment b) Gentle-Gerlach experiment c) quartz-crystald) nuclear magnetic resonance machine e) atomic clock

014 qmult 01800 1 1 1 easy memory: second defined.

92 Chapt. 14 Spin

22. The spin state energy level separation of a 133Ce atom used in an atomic clock to define thesecond corresponds by definition to a frequency of:

a) 9 192 631 770 Hz.b) 3.141 592 65 Hz.c) 2.718 28 Hz.d) 0.577 215 66 Hz.e) 299 792 458 Hz.


014 qfull 00100 2 5 0 moderate thinking: Pauli matrices in detail1. The Pauli spin matrices are

σx =

(

0 11 0

)

, σy =

(

0 −ii 0

)

, and σz =

(

1 00 −1

)

.

a) Are the Pauli matrices Hermitian?

b) What is the result when Pauli matrices act on general vector

(

ab

)

?

c) Diagonalize the Pauli matrices: i.e., solve for their eigenvalues and NORMALIZEDeigenvectors. NOTE: The verb ‘diagonalize’ takes its name from the fact that a matrixtransformed to the representation of its own eigenvectors is diagonal with the eigenvaluesbeing the diagonal elements. One often doesn’t actually write the diagonal matrix explicitly.

d) Prove thatσiσj = δij1 + iεijkσk ,

where i, j, and k stand for any of x, y, and z, 1 is the unit matrix (which can often beleft as understood), δij is the Kronecker delta, εijk is the Levi-Civita symbol, and Einsteinsummation is used. HINT: I rather think by exhaustion is the only way: i.e., extremetiredness.

e) Prove[σi, σj ] = 2iεijkσk and σi, σj = 2δij ,

where σi, σj = σiσj + σjσi is the anticommutator of Pauli matrices. HINT: You shouldmake use of the part (d) expression.

f) Show that a general 2× 2 matrix can be expanded in the Pauli spin matrices plus the unitmatrix: i.e.,

(

a bc d

)

= α1 + ~β · ~σ ,

where ~σ = (σx, σy, σz) is the vector of the Pauli matrices. HINT: Find expressions for theexpansion coefficients α, βx, βy, and βz.

g) Let ~A and ~B be vectors of operators in general and let the components of ~B commute withthe Pauli matrices. Prove

( ~A · ~σ)( ~B · ~σ) = ~A · ~B + i( ~A× ~B) · ~σ .

Chapt. 14 Spin 93

HINT: Make use of the part (d) expression.

014 qfull 00110 1 3 0 easy math: diagonalization of y Pauli spin matrixExtra keywords: (CDL-203:2), but it corresponds to only part of that problem

2. The y-component Pauli matrix (just the y-spin matrix sans the h−/2 factor) expressed in termsof the z-component orthonormal basis (i.e., the standard z-basis with eigenvectors |+〉 and |−〉)is:

σy =

(

0 −ii 0

)

.

Diagonalize this matrix: i.e., solve for its eigenvalues and NORMALIZED eigenvectors writtenin terms of the standard z-basis eigenvector kets or, if your prefer, in column vector form forthe z-basis. One doesn’t have to literally do the basis transformation of the matrix to thediagonal form since, if one has the eigenvalues, one already knows what that form is. Inquantum mechanics, literally doing the diagonalization of the matrix is often not intended by adiagonalization.

014 qfull 00200 2 3 0 mod math: spin 1/2, spin Sx + Sy

Extra keywords: (Ga-241:9), spin 1/2, spin Sx + Sy, diagonalization3. Consider a spin 1/2 system. Find the eigenvectors and eigenvalues for operator Sx + Sy. Say

the system is in one of the eigen-states for this operator. What are the probabilities that an Sz

measurement will give h−/2?

014 qfull 00250 2 5 0 moderate thinking: Euler formula for matricesExtra keywords: Reference Ba-306, but I’ve generalized the result

4. Say that A is any matrix with the property that A2 = 1, where 1 is the unit matrix. If wedefine eixA by

eixA =

∞∑

ℓ=0

(ixA)ℓ

ℓ!

(where x is a scalar), show that

eixA = 1 cos(x) + iA sin(x) .

This last expression is a generalization of Euler’s formula (Ar-299).

014 qfull 00300 3 5 0 tough thinking: rotation parametersExtra keywords: (Ba-330:1b), but there is much more to this problem

5. The unitary spin 1/2 rotation operator is

U(~α) = e−i~S·~α/h− = e−i~σ·~α/2 ,

where ~α is the vector rotation angle: ~α = αα with α being the angle of a right-hand rule rotationabout the axis aligned by α. To rotate the spin component operator Sz from the z direction tothe n direction one uses

~S · n = U(~α)SzU(−~α)

(Ba-305) To rotate a z basis eigenstate into an n basis eigenstate one uses

|n±〉 = U(~α)|z±〉

(Ba-306).

a) Expand U(~α) into an explicit 2 × 2 matrix that can be used directly. Let the componentsof α be written αx, etc.

94 Chapt. 14 Spin

b) Now write out U(~α)|z±〉 explicitly.

c) You are given a general normalized spin vector

|n+〉 =

(

a+ ibc+ id

)

.

Find expressions for α and ~α that yield this vector following a rotation of |z+〉. Then forthose α and ~α written in terms of a, b, c, and d, find the rotated |z−〉 state |n−〉. Showexplicitly that |n−〉 is normalized and orthogonal to |n+〉.

d) We gone so far: why quit now. Using our explicit matrix version of U(~α) find explicitexpressions for the components of n for a rotation from z in terms of α and the componentsof α. One has to solve for the components from

~σ · n = U(~α)σzU(−~α) .

HINT: Write U(~α) in simplified symbols until it’s convenient to switch back to the propervariables: e.g.,

U(~α) =

(

a+ ib −c+ idc+ id a− ib

)

,

where a, b, c, and d have the same meanings as you should have found in the part (c)answer.

014 qfull 00400 1 5 0 easy thinking: electron spin in B-field HamiltonianExtra keywords: electron spin in magnetic field Hamiltonian

6. What is the Hamiltonian fragment (piece, part) that describes the energy of an electron spinmagnetic moment in a magnetic field? This fragment in a Schrodinger equation can sometimesbe separated from the rest of the equation and solved as separate eigenvalue problem. Solve thisseparated problem. The intrinsic angular momentum operator is ~S and assume the magneticfield points in the z direction. HINTS: Think of the classical energy of a magnetic dipole in amagnetic field and use the correspondence principle. This is not a long question.

014 qfull 00500 2 5 0 moderate thinking: classical Larmor precession7. Let’s tackle the classical Larmor precession.

a) What is Newton’s 2nd law in rotational form?

b) What is the torque on a magnetic dipole moment ~µ in a magnetic field ~B? HINT: Anyfirst-year text will tell you.

c) Say that the magnetic moment of a system is given by ~µ = γ~L, where γ is the gyromagnetic

ratio and ~L is the system’s angular momemtum. Say also that there is a magnetic field~B = (0, 0, Bz). Solve for the time evolution of ~L using Newton’s 2nd law in rotational form

assuming the INITIAL CONDITION ~L(t = 0) = (Lx,0, 0, Lz,0). HINTS: You should

get coupled differential equations for two components of ~L. They are not so hard to solve.For niceness you should define an appropriate Larmor frequency ω.

014 qfull 00600 3 5 0 tough thinking: quantum mech. Larmor spin precessionExtra keywords: (Ba-330:1a), but there is much more to this problem

8. Consider a spin 1/2 particle with magnetic moment ~M = γ~S. We put a uniform magnetic

field in z direction: thus ~B = Bz z. As usual we take the z-basis as the standard basis for theproblem.

a) Determine the normalized eigenstates for the Sx, Sy, and Sz operators in the z-basis. What arethe eigenvalues?

Chapt. 14 Spin 95

b) Now expand the eigenvectors for Sz in the bases for Sx and Sy. You will need the expansionsbelow.

c) If we consider only the spin degree of freedom, the Hamiltonian for the system is

H = −~m · ~B = −γ~S · ~B ,

where γ is a constant that could be negative or positive. Sometimes γ is called the gyromagneticratio (CDL-389), but the expression gyromagnetic ratio is also used for the Lande g factor whichitself has multiple related meanings. What are the eigenvalues and eigenvectors of H in thepresent case? HINT: Defining an appropriate Larmor frequency ω would be a boon furtheron.

d) The time-dependent Schrodinger equation in general is

ih− ∂

∂t|Ψ〉 = H |Ψ〉 .

What is the formal solution for |Ψ(t)〉 in terms of H and a given |Ψ(0)〉. HINT: Expand |Ψ(0)〉in a the eigenstates of H which you are allowed to assume you know.

e) For our system you are given

|Ψ(0)〉 = a+|z+〉+ a−|z−〉 .

What is |Ψ(t)〉? What are the probabilities for measuring spin up and down in the z directionand what is 〈Sz〉?

f) What are the probabilities for measuring spin up and down in the x direction and what is 〈Sx〉?Try to get nice looking expressions.

g) What are the probabilities for measuring spin up and down in the y direction and what is 〈Sy〉?Try to get nice looking expressions.

h) What can you say about the vector of spin expectation values given the answers to parts (f)and (g)?

i) Now given the initial state as |x+〉, what are 〈Sx〉, 〈Sy〉, and 〈Sz〉 in this special case?

014 qfull 00700 2 5 0 tough thinking: spin algebra generalizedExtra keywords: (Ba-331:3)

9. Spin algebra can be used usefully for situations not involving spin. Say we have an atomor molecule with two isolated stationary states: i.e., there can be perturbation coupling andtransitions between the two states, but no coupling to or transitions to or from anywhere else.Let the states be |+〉 and |−〉 with unperturbed energies ǫ+ and ǫ−; let ǫ+ ≥ ǫ−. The statesare orthonormal.

a) Write the Hamiltonian for the states in matrix form and then decompose it into a linearcombination of Pauli spin matrices and the unit matrix. What are the eigenstates |+〉 and|−〉 in column vector form? Note there is no spin necessarily in this problem: we are justusing the Pauli matrices and all the tricks we have learned with them.

b) Now we add a perturbation electric field in in the z direction: E = E0 cos(ωt). You aregiven that the diagonal elements of the dipole moment matrix are zero and that the offdiagonal elements are both equal to the real constant µ: i.e., µ = 〈+|ez|−〉 = 〈−|ez|+〉.Note µ can be positive or negative. Write down the Hamiltonian now.

c) Write the Schrodinger equation for the perturbed system and then make a transformationthat eliminates the unit matrix term from the problem. Do we ever really need to transformthe state expressions back?

96 Chapt. 14 Spin

d) Show that a pretty explicit, approximate solution for the (transformed) Schrodingerequation is

|Ψ〉 = e−iωtσz/2e−iΩσt/2|Ψ(0)〉 ,

where

Ω =√

(ω0 − ω)2 + ω21 and σ =

(ω0 − ω)

Ωσz − ω1

Ωσx .

The solution is valid near resonance: i.e., the case of ω ≈ ω0, where ω0 ≡ (ǫ+ − ǫ−)/h−.In order to get the solution we have averaged over times long enough to eliminate some ofthe high frequency behavior. To do this one assumes that |ω1| = |µE0/h−| << |ω| ≈ |ω0|.HINT: The problem is pretty much isomorphic to the spin magnetic resonance problem.

e) Given that the initial state is |+〉, what are the probabilities that the system is in |+〉 and|−〉 at any later time? What are corresponding probabilities if the initial state is |−〉? Doany of these probabilities have high frequency behavior: i.e., time variation with frequencyof order ω ≈ ω0 or greater?

f) The factor e−iωtσz/2 in the solution

|Ψ〉 = e−iωtσz/2e−iΩσt/2|Ψ(0)〉

is actually physically insufficient to give the high frequency behavior—although it is rightin itself—since we dropped some high frequency behavior in deriving the solution. Thusany high frequency behavior predicted by the solution can’t be physically accurate. Youshould have found in part (e) that the high frequency behavior from e−iωtσz/2 canceled outof the probability expressions. Is there any reason for keeping the factor e−iωtσz/2 in theformal solution? If there is a reason, what is it? HINT: There is a reason.

Chapt. 15 Time-Independent Approximation Methods


015 qmult 00090 1 1 1 easy memory: quantum perturbation SantaExtra keywords: the Christmas question

1. Santa Claus discovers that an intractable time-independent Schrodinger equation (i.e., aHamiltonian eigen-problem) has an approximated form that is exactly solvable and has solutionsthat must be nearly those of the original problem. The approximated form eigen-solutions arealso NOT degenerate. Not being a sage for nothing, Santa leaps to the conclusion that theoriginal problem can now be solved by:

a) time-independent perturbation theory.b) checking it twice.c) getting the elves to work on it.d) unthrottling the antlers,

bidding Blixen to bound to the world’s height,and chasing the dim stars,pass into nightness and out of all sight.

e) peace on Earth and goodwill to humankind.

015 qmult 00100 1 1 1 easy memory: time-independent perturbation2. Time-independent weak-coupling non-degenerate perturbation theory assumes that the

stationary states and eigen-energies of a system can be expanded in convergent power series ina perturbation parameter about, respectively:

a) the stationary states and eigen-energies of a solvable system.b) the eigen-energies and stationary states of an unsolvable system. c) the origin.d) the center. e) infinity.

015 qmult 00200 1 1 5 easy memory: zeroth order perturbation3. The zeroth order perturbation of a system is:

a) the most strongly perturbed system. b) the mostest strongly perturbed system.c) the deeply disturbed system. d) the negatively perturbed systeme) the unperturbed system.

015 qmult 00300 1 1 2 easy memory: 1st order energy correction4. The formula

E1stn = E(0)

n + λ〈ψ(0)n |H(1)|ψ(0)

n 〉is:

a) the eigen-energy of eigen-state n to 0th order in perturbation λH(1).b) the eigen-energy of eigen-state n to 1st order in perturbation λH(1).c) the energy of eigen-state n to 2nd order in perturbation λH(1).d) the eigen-state n to 1st order in perturbation λH(1).e) the eigen-state n to 2nd order in perturbation λH(1).

015 qmult 00400 1 4 4 easy deducto-memory: 1st order eigen state correction

97

98 Chapt. 15 Time-Independent Approximation Methods

5. The formula

|ψ1stn 〉 = |ψ(0)

n 〉 + λ∑

all k, except k 6=n

〈ψ(0)k |H(1)|ψ(0)

n 〉E

(0)n − E

(0)k

|ψ(0)k 〉

is:

a) the eigen-energy of eigen-state n to 0th order in perturbation λH(1).b) the eigen-energy of eigen-state n to 1st order in perturbation λH(1).c) the energy of eigen-state n to 2nd order in perturbation λH(1).d) the eigen-state n to 1st order in perturbation λH(1).e) the eigen-state n to 2nd order in perturbation λH(1).

015 qmult 00500 1 1 3 easy memory: 2nd order energy correction6. The formula

E2ndn = E(0)

n + λ〈ψ(0)n |H(1)|ψ(0)

n 〉 + λ2∑

all k, except k 6=n

|〈ψ(0)k |H(1)|ψ(0)

n 〉|2

E(0)n − E

(0)k

is:

a) the eigen-energy of eigen-state n to 0th order in perturbation λH(1).b) the eigen-energy of eigen-state n to 1st order in perturbation λH(1).c) the eigen-energy of eigen-state n to 2nd order in perturbation λH(1).d) eigen-state n to 1st order in perturbation λH(1).e) eigen-state n to 2nd order in perturbation λH(1).

015 qmult 00600 1 4 1 easy deducto-memory: degeneracy and perturbation7. “Let’s play Jeopardy! For $100, the answer is: A common cause for the obvious failure of

time-independent weak-coupling perturbation theory.”

What is , Alex?

a) degeneracy b) tarnation c) subversion d) lunacy e) regency

015 qmult 01000 1 4 5 easy deducto-memory: diagonalizationExtra keywords: mathematical physics

8. “Let’s play Jeopardy! For $100, the answer is: A standard, non-perturbative approximatemethod of solving for the eigen-energies and stationary states of a system. If the system is in afinite Hilbert space (i.e., a finite function space), the method can be done for an exact solution.”

What is , Alex?

a) perturbation theory b) divagation c) strangulation d) triangulatione) diagonalization

015 qmult 01020 1 1 2 easy memory: 2x2 eigenvalues9. The values

E =1

2

[

(H11 +H22) ±√

(H11 −H22)2 + 4|H12|2]

are:

a) the stationary states of a 2 × 2 Hamiltonian matrix.b) the eigen-energies of a 2 × 2 Hamiltonian matrix.c) the eigen-energies of a 3 × 3 Hamiltonian matrix.d) the stationary states of a 3 × 3 Hamiltonian matrix.e) the 1st order non-degenerate perturbation correction energies.

Chapt. 15 Time-Independent Approximation Methods 99


015 qfull 00100 2 5 0 moderate thinking: what is a perturbation?1. What is a perturbation?

015 qfull 00200 2 5 0 moderate thinking: basic perturbation hypothesis2. What is the basic non-degenerate perturbation method hypothesis?

015 qfull 00300 2 5 0 moderate thinking: smallness parameter3. What is role of the smallness parameter in non-degenerate perturbation theory?

015 qfull 00400 2 5 0 moderate thinking: 2nd bigger than 1st4. If all the 2nd order non-degenerate perturbation corrections are greater than the 1st order ones,

what might you suspect?

015 qfull 00500 2 5 0 moderate thinking: 2nd bigger than 1st all zero5. If all the 2nd order non-degenerate perturbation corrections are greater than the 1st order ones,

but the 1st order ones were all identically zero, what might you suspect?

015 qfull 01400 2 3 0 moderate math: infinite square well Dirac delta perturbation 1Extra keywords: (Gr-225:6.1) Dirac delta perturbation, 1-dimensional infinite square well

6. Say you have a 1-dimensional infinite square well with

V (x) =

0 for the x range 0 to a;∞ otherwise.

a) Solve for the eigen-states (i.e., stationary states) and eigen-energies from the time-independent Schrodinger equation. You must properly normalize the eigen-states statesto answer part (b) correctly.

b) Say we add the Dirac delta function perturbation Hamiltonian

H(1) = cδ(x− a/2) .

What is the general expression for this perturbation for the first order perturbation energycorrection for all eigen-states?



V (x) =


The stationary states are

ψn(x) =

√

2

asin (knx) ,

where

kna = nπ and kn =nπ

a

with n = 1, 2, 3, . . . as allowed quantum numbers. The eigen-energies are

En =h−2

2m

(π

a

)2

n2 .


Say we add the Dirac delta function perturbation Hamiltonian

H(1) = cδ(x− a/2) .

What is the general formula for this perturbation for the first order perturbation energycorrection for all eigen-states? Simplify the formula as much as possible.



V (x) =


The stationary states are

ψn(x) =

√

2

asin (knx) ,

wherekna = nπ and kn =

nπ

a


En =h−2

2m

(π

a

)2

n2 .

We now add the Dirac delta function perturbation Hamiltonian

H(1) = cδ(x− a/2) .

a) Can we use non-degenerate perturbation theory for the infinite square well? Why or why not?

b) What is the general formula for the perturbation for the 1st order perturbation energy correctionfor all eigen-states? Simplify the formula as much as possible.

c) Now evaluate a general matrix element for the perturbation

〈ψm|H(1)|ψn〉 .

d) Simplify the general matrix element by inventing a simple function quantum number ℓ that iszero for ℓ even and 1 for ℓ odd and another simple function for odd quantum number ℓ that is 1for (ℓ− 1)/2 even and −1 for (ℓ− 1)/2 odd. HINT: If you this can’t get this part, go on sincethe latter parts don’t require it.

e) Write out the 1st order perturbation correction formula for a general state n in as explicit andas simplified a form as reasonably possible. Note the correction is wanted, not the full 2ndorder corrected state. HINT: Just leave the unperturbed states in the ket form |ψ0

n〉. Forcompactness, one doesn’t want to be explicit about them.

f) Write out the 2nd order perturbation correction formula for a general eigen-energy n in asexplicit and as simplified a form as reasonably possible. Note the correction is wanted, not thefull 2nd order corrected eigen-energy.

014 qfull 01410 2 3 0 moderate math: 2-particle Dirac delta perturbation 19. You are give a complete set of orthonormal 1-dimensional single-particle states ψn(x), wheren is the indexing quantum number that determines energy. The state are NOT degenerate in


energy. NOTE: Some parts of this problem can be done independently. So don’t stop at anypart that you can’t immediately solve.

a) Now we need a general way to evaluate the eigen-energies for a symmetrized multi-particlestate for a set of non-interacting identicle particles. The single-particle Hamiltonian Hi

acting on single-particle state ψn(xi) gives

Hiψn(xi) = Enψn(xi) .

A multi-particle state can be created from the single particle states. Each particle for themulti-particle state has its own single-particle Hamiltonian Hi, but these Hamiltoniansare have identical formulae. For multi-particle state of I particles, the multi-particleHamiltonian is

HI =

I∑

i=1

Hi .

A simple product state (i.e., an unsymmetrized state) for I particles) is

ψI,prod = ψn1(x1)ψn2

(x2) . . . ψnI(xI) .

This state though unphysical (since unsymmetrized) is an eigenstate of the multi-particleHamiltonian. We find

HIψI,prod =

I∑

j=1

Enj

ψI,prod ,

and so the eigen-energy of the simple product state is

EI =

I∑

i=1

Eni,

where the sum is over the set of n indexes that are in the simple product state.What is the eigen-energy of the symmetrized state that is constructed from the simple

product state? HINT: In symmetrizing the multi-particle state, all I! permutations of thecoordinate labels xi occur among the product states, but the set of n indexes does notchange. Each ni just occurs once in each term of the symmetrized state. The answer is byinspection.

b) Say you have two non-interacting identical particles that can occupy the set of states. Theycould be bosons or fermions. If the particles are bosons, they are spinless. If the particlesare fermions, they are in the same spin state. Thus, for both particles, we don’t have toconsider spin any further in this problem.

Give the general normalized symmetrized wave function for the two particles. Labelthe single-particle states m and n and the coordinates for the particles x1 and x2. Be surethe normalization is correct for both the m 6= n and m = n cases.

c) Now say we turn on a perturbation Hamiltonian

H(1) = aV0δ(x1 − x2) ,

where a is a characteristic length, V0 is strength factor, and δ(x1 − x2) is a Dirac deltafunction. The perturbation gives an interaction between the two particles. Determine thefirst order non-degenerate perturbation energy correction for the general two-particle state.Be as explicit as possible. HINT: For 2-dimensional space,

〈φ|Q|ψ〉 =

∫ ∞

−∞

∫ ∞

−∞

φ(x1, x2)∗Qψ(x1, x2) dx1 dx2 ,


where |φ〉 and |ψ〉 are general states and Q is a general quantum mechanical observable(i.e., Hermitian operator).

d) Let’s now specialize to the infinite square complete set of states. Recall the potential forthe infinite square well is

V (x) =


The single-particle stationary states of this set are

ψn(x) =

√

2

asin (knx) ,

wherekna = nπ and kn =

nπ

a


En =h−2

2m

(π

a

)2

n2 .

What is the energy for a general two-particle infinite-square-well state withoutperturbation? HINT: Remember the part (a) answer.

e) Determine the 1st order non-degenerate perturbation energy correction for the general two-particle infinite-square-well state. Be as explicit as possible. HINT: You will need a coupleof trig identities:

sin2(A) =1

2[1 − cos(2A)] and cos(A) cos(B) =

1

2[cos(A−B) + cos(A+B)] .

Just keep going step by step carefully.

f) Using the part (e) answer, what are the 1st order non-degenerate perturbation energycorrections for the case of m = n and the case of m 6= n.

015 qfull 01412 3 5 0 tough thinking: 2-particle Dirac delta perturbation 2Extra keywords: (Gr-226:6.3)

10. The single-particle stationary states and eigen-energies for a 1-dimensional infinite square wellfor region [0, a] are, respectively,

ψn(x) =

√

2

asin(nπ

ax)

and En =h−2

2m

(π

a

)2

n2 .

a) What is the expression for elementary 2-particle stationary states for NON-identicalparticles of the same mass? (Label the particles a and b for convenience and assumethe particles are spinless. Label the states n and n′ for convenience too.) What is thegeneral expression for the energy of such 2-particle states? What are all the possiblereduced energies (i.e., n2 + n′2) up n = n′ = 7? These energies can be called energy levels:the levels may correspond to more than one state. (You are permitted to use a computerprogram to generate these.) Are there any degeneracies with these energies? Rememberthe particles are not identical.

b) Now suppose we turn on a perturbation potential for the non-identical particles of the form

H(1) = V (xa, xb) = aV0δ(xa − xb) .


What is the expression for the diagonal matrix element

H(nn′)(nn′) = 〈ψnn′(xa, xb)|H(1)|ψnn′(xa, xb)〉 .

If you expand sin θ in exponentials evaluating, the matrix element is pretty easy, but you dohave to treat the cases where n 6= n′ and n = n′ a bit differently. Given the diagonal matrixelements can you do (weak-coupling) perturbation theory on all the 2-particle states?

c) What is the expression for elementary 2-particle stationary states for identical spinlessbosons? (Label the particles a and b for convenience. Note we have turned off theperturbation.) What is the general expression for the energy of such 2-particle states?What are all the possible reduced energies (i.e., n2 + n′2) up n = n′ = 7? (You don’t haveto do part (a) all over again, just mutatis mutandis it.) Are there any degeneracies withthese energies? Remember the particles are identical.

d) Now suppose we turn on a perturbation potential of part (b) for the identical bosons. Whatis the expression for the diagonal matrix element


If you expand sin θ in exponentials evaluating, the matrix element is pretty easy, but youdo have to treat the cases where n 6= n′ and n = n′ a bit differently. Note the perturbationcorrection is a bit different from the non-identical particle case. Why? Given the matrixelements can you do (weak-coupling) perturbation theory on all the 2-particle states?

e) What is the expression for elementary 2-particle stationary states for identical fermionswhen we assume spin coordinates are identical. Since the spin coordinates are identical,the spin part of the single-particle states are symmetrical. Don’t bother writing downspinors or such. (Label the particles a and b for convenience. Note we have turned offthe perturbation.) What is the general expression for the energy of such 2-particle states?What are all the possible reduced energies (i.e., n2 + n′2) up n = n′ = 7? (You don’t haveto do part (a) all over again, just mutatis mutandis it.) Are there any degeneracies withthese energies? Remember the particles are identical.

f) Now suppose we turn on a perturbation potential of part (b) for the identical fermions.What is the expression for the diagonal matrix element


Don’t whine: this is easy if you see the trick. Why do you get the simple result youget? Given the matrix element, can you do (weak-coupling) perturbation theory on all the2-particle states?

g) What does the Dirac delta potential

V (xa − xb) = aV0(xa − xb)

imply or do physically?

015 qfull 01500 1 3 0 easy math: SHO 1st order perturbation cxExtra keywords: SHO 1st order perturbation cx

11. Say you add a perturbation potential cx to a 1-dimensional simple harmonic oscillator (SHO)system. Calculate all the first order weak-coupling perturbation corrections for the eigen-energies. Recall the 1st order perturbation energy correction is given by

E(1)n = 〈ψ(0)

n |H(1)|ψ(0)n 〉 = 0 ,


where the |ψ0n〉 are unperturbed eigenstates. HINT: Think about the parity of SHO energy

eigenstates.

015 qfull 01600 2 3 0 mod math: SHO exact cx perturbationExtra keywords: (Gr-227:6.5), SHO, linear perturbation cx, exact cx solution

12. Say you added a perturbation H(1) = cx to the 1-dimensional simple harmonic oscillator (SHO)Hamiltonian, and so have

H =p2

2m+

1

2mw2x2 + cx

for the Hamiltonian. An exact solution to the time independent Schrodinger equation is, infact, possible and easy since the new problem is still a SHO problem.

a) Let’s consider just the mathematical aspects of the problem first. Given a quadratic

y = ax2

with a > 0, where is its minimum and roots? Say you now add bx to get

y = ax2 + bx .

Where are the minimum and roots now? By measuring the horizontal coordinate from a neworigin it is possible to eliminate the linear dependence on the horizontal coordinate. Find thisnew origin. From a geometrical point of view what have you done by adding bx to y = ax2: i.e.,what has happened to the parabola on the plane? Sketch a plot of the original and translatedparabolae and the curve y = bx. Why is should it be clear that adding the linear term bx thatthe mininum of the curve will be shifted downward?

b) Now that the math is clear what about the physics. What are the classical forces associatedwith the potentials

1

2mω2x2 , cx , and

1

2mω2x2 + cx ?

What are the equilibrium points of the forces? What are the potential energies of the first andthird equilibrium points? What has adding the cx potential done to the potential well of theSHO? How could you reduce the problem with the third potential to that with the first?

c) Now reduce time independent Schrodinger problem with the given Hamiltonian to the SHOproblem. What are the solutions in terms of horizontal coordinate distance from the new originand what are eigen-energies of the solutions? (I don’t mean solve for the solutions. Just whatare known solutions for reduced problem.)

015 qfull 01700 3 3 0 mod math: SHO and 2nd order perturbation cxExtra keywords: SHO and 2nd order perturbation cx

13. Say you add a perturbation potential cx to a 1-dimensional simple harmonic oscillator (SHO)system. Give the formula for the 2nd order weak coupling perturbation correction for thisspecial case simplified as much as possible. HINT: You will probably find the following matrixelement formula for SHO eigenvectors useful:

〈Ψk|x|Ψn〉 =

1

β

√

max(k, n)

2if |k − n| = 1;

0 otherwise,

where β =√

mω/h− (e.g., Mo-406).

015 qfull 03000 3 3 0 tough math: SHO and 2nd order x3 pre-perturbationExtra keywords: SHO and 2nd order x3 pre-perturbation


14. In preparation for calculating the 1st order perturbation wave function correction and the 2ndorder perturbation energy correction for the 1-dimensional simple harmonic oscillator (SHO)system with perturbation potential cx3, one needs to find a general expression for

〈ψk|x3|ψn〉 .

Find this expression simplified as much as possible.

INSTRUCTIONS: You will need the following to formulae (which I hope are correct)

1√2β

[√n+ 1ψn+1(x) +

√nψn−1(x)

]

= xψn(x)

and

〈ψk|x2|ψn〉 =

2n+ 1

2β2if k = n;

√

[max(k, n) − 1] max(k, n)

2β2if |k − n| = 2;

0 otherwise,

where β =√

mω/h−. There are seven initial cases (one being zero) to find and five final cases

after combining initial cases with same k and n relation. Write the expressions in terms of n,not k. You will simply have to work carefully and systematically to grind out the cases. Whatis the appropriate Kronecker delta function to go with each case so that one can put themin a sum over k in the 2nd order perturbation formulae? Make the Kronecker deltas in theform δk,f(n) where f(n) is an expression like, e.g., n− 1. Since k in the sum for the 2nd orderperturbation runs only from zero to infinity is there any special treatment needed for includingcases with Kronecker deltas like δk,n−1 for n = 0? HINT: Are such cases ever non-zero whenthey should be omitted?

015 qfull 03100 3 3 0 mod math: SHO and 2nd order cx3 perturbationExtra keywords: SHO and 2nd order cx3 perturbation

15. The following result is for simple harmonic oscillator eigenvectors:

〈ψk|x3|ψn〉 =1

2√

2β3

3(n+ 1)√n+ 1 if k = n+ 1 with δk,n+1;

3n√n if k = n− 1 with δk,n−1;

√

(n+ 1)(n+ 2)(n+ 3) if k = n+ 3 with δk,n+3;√

(n− 2)(n− 1)n if k = n− 3 with δk,n−3;

0 otherwise.

Using this expression find the general expression for the SHO for the 2nd order weak-couplingperturbation corrections to the eigenstate energies for a perturbation potential cx3. Why canyou use the expression above without worrying about the fact that sum over states from zeroto infinity doesn’t include states with index less than zero.

015 qfull 03110 2 5 0 moderate thinking: 4x4 eigenproblem/perturbation16. You are given a zeroth order Hamiltonian matrix

H(0) =

1 0 0 00 1 0 00 0 1 00 0 0 −1

.


a) Solve for the eigenvalues and normalized eigenvectors by inspection. You should label thestates 1, 2, 3, and 4 for convenience. Is there any degeneracy and if so what are thedegenerate states?

b) The evil wizard of physics now turns on a perturbation and the Hamiltonian becomes

H =

1 ǫ 0 0ǫ 1 0 00 0 1 ǫ0 0 ǫ −1

,

where ǫ is a small quantity. Solve for the exact eigenvalues and normalized eigenvectorsin this case: i.e., diagonalize the perturbed Hamiltonian matrix. Is there any degeneracynow? HINT: Is there any reason why the two 2×2 blocks in the matrix cannot be treatedas separate eigenvalue problems and the two-component eigenvectors extended trivially forthe 4 × 4 problem?

c) Do weak-coupling non-degenerate perturbation theory to solve for the energy to 2nd orderfor those initial eigenstates which are NOT degenerate. HINT: All the perturbationmatrix elements can be found in the part (b) question.

015 qfull 03300 3 5 0 tough thinking: perturbation and variationExtra keywords: (Gr-235:6.9)

17. Consider quantum system of 3 dimensions with initial Hamiltonian

H(0) =

1 0 00 1 00 0 2

and perturbed Hamiltonian

H =

1 − ǫ 0 00 1 ǫ0 ǫ 2

.

Note we assume ǫ << 1. Also note that H(0) andH are matrix Hamiltonians: i.e., Hamiltonians

in a particular representation. The matrix elements are 〈φi|H(0)op |φk〉. 〈φi|Hop|φk〉, respectively,

where H(0)op and Hop are operator versions of the Hamiltonian and |φi〉 are some orthonormal

basis. Usually we drop the “op” subscript and allow context to tell whether the Hamiltonian isin matrix or operator representation.

a) Solve by inspection for the eigen-energies and eigenvectors of the initial unperturbedHamiltonian. To help with the rest of the problem label the states 1, 2, and 3 in some sensibleorder.

b) Solve for the exact eigen-energies and normalized eigenvectors of the perturbed Hamiltonian:i.e., diagonalize the perturbed Hamiltonian matrix. HINTS: It’s not so hard—if you don’tmake a mistake in the first step.

c) Expand the exact eigen-energies and eigenvectors (where applicable) to 2nd order in small ǫ.(Note I mean Taylor expansion, not perturbation series expansion although the two expansionare closely related in this case.) Simplify the eigenvectors to nice forms so that it is easy to seewhich perturbed vector grew out of which unperturbed vector as ǫ grew from 0.

d) Determine from (weak-coupling) perturbation theory the energies to 2nd order and theeigenvectors to 1st order of the perturbed Hamiltonian. How do these results compare withthose of the part (c) answer? HINT: Perturbation theory can be applied to the degeneratestates in this case because they are completely uncoupled.


e) Now use the truncated Hamiltonian matrix method (or linear variational method if you knowit) to find approximate eigen-energies and eigenvectors for the two initially degenerate eigen-energy states. To what order goodness in small ǫ are the results? Why the are results for oneperturbed state exact and for the other rather poor compared to the exact results?

015 qfull 03400 2 3 0 moderate math: variational x**4 potential18. You are given a 1-dimensional Hamiltonian with a quartic potential:

H = − h−2

2m

∂2

∂x2+ V0

(x

a

)4

,

where V0 is a constant. The Hamiltonian applies over the whole x-axis.

a) Write H in a dimensionless form in units of energy h−2/(2ma2), with y = x/a, and with a

dimensionless potential constant V1.

b) Show definitively that a trial Gaussian wave function

ψ(x) = Ae−βx2/a2

,

where β is a variational parameter, cannot be an eigenfunction of the Hamiltonian for anyvalue of β. Remember a trial wave function could fortuitously have the right form to bean eigenfunction.

c) Write down the dimensionless variational energy ǫv using the trial Gaussian wave functionand solve for ǫv as an explicit function of β and V1. HINT: Remember to account fornormalization.

d) Sketch ǫv as a function of β on a schematic plot.

e) Determine the βmin value that makes ǫv stationary and, in fact, a minimum. What isminimum ǫv,min? Why can’t ǫv,min be the true ground state energy of the dimensionlessHamiltonian? What is the qualitative relation between ǫv,min and ǫground.

Chapt. 16 Variational Principle and Variational Methods


016 qmult 00700 1 1 3 easy memory: equivalent postulates1. If two postulates are said to be equivalent, then

a) one can be derived from the other, but not the other from the one.b) the other can be derived from the one, but not the one from the other.c) each one can be derived from the other.d) neither can be true.e) both must be true.

016 qmult 00800 1 4 5 easy deducto-memory: variational principle2. “Let’s play Jeopardy! For $100, the answer is: Usually the demand that an action (or action

integral) be stationary with respect to arbitrary variation in a function appearing somehow inthe integrand.”

a) What is a Hermitian conjugate, Alex?b) What is an unperturbation principle, Alex?c) What is a perturbation principle, Alex?d) What is an invariation principle, Alex?e) What is a variational principle, Alex?

016 qmult 00900 1 1 3 easy memory: quantum mechanics action3. In non-relativistic quantum mechanics the action of the usual variation principle is:

a) the integral of angular momentum.b) the derivative of angular momentum.c) the expectation value of the Hamiltonian.d) the time independent Schrodinger equation.e) the Dirac equation.

016 qmult 01000 1 1 1 easy memory: stationary action4. An exact solution |φ〉 to the time-independent Schrodinger equation is the one that by the

variational principle in quantum mechanics makes the action

E(φ) =〈φ|H |φ〉〈φ|φ〉

be stationary with respect to:

a) arbitrary variations of the state |φ〉 (i.e., δE(φ) = 0).b) some variations of the state |φ〉.c) no variations of the state |φ〉.d) reasonable variations of the state |φ〉.e) unreasonable variations of the state |φ〉.

108

Chapt. 16 Variational Principle and Variational Methods 109

016 qmult 01100 1 1 5 easy memory: simple variational method5. In the simple variational method one takes a parameterized trial wave function and finds the

parameters that make the expectation value of the Hamiltonian:

a) a maximum.b) 1.c) negative.d) positive.e) a minimum.

016 qmult 01200 1 4 3 easy deducto-memory: linear variation method6. “Let’s play Jeopardy! For $100, the answer is: The justification for the linear variational method

(or Rayleigh-Ritz method or truncated Hamiltonian matrix eigen-problem).”

a) What is Hermitian conjugation, Alex?b) What is bra/ket notation, Alex?c) What is the quantum mechanics variational principle, Alex?d) What is the Dirac principle, Alex?e) What is the cosmological principle, Alex?

016 qmult 01500 1 4 1 easy deducto-memory: repulsion of the energy levels7. Any perturbation applied to a two-level system that is initially degenerate causes:

a) a repulsion of the energy levels.b) an attraction of the energy levels.c) a warm and affectionate relationship between the energy levels.d) a wonderful, meaningful togetherness of the energy levels.e) an eternal soul-bliss of the energy levels.


016 qfull 00010 1 5 0 easy thinking: equivalent results1. If two different looking theorems or postulates were said to be equivalent what would that

mean?

016 qfull 00020 2 5 0 moderate thinking: variational principle and method2. Are the variational principle and the variational method the same thing? Explain please.

016 qfull 00030 1 5 0 easy thinking: what is a stationary point?3. What does it mean to say a function is stationary at a point?

016 qfull 00040 2 3 0 moderate math: differentiation for stationarity4. Take the derivative of

E(α) =5

4

h−2

mα2+

1

14mω2α2

and determine the stationary point. Just by imagining the function’s behavior in the largeand small α limits determine whether the stationary point is a minimum. Give the analyticexpression for E(α) at the stationary point.

016 qfull 00050 2 5 0 moderate thinking: Snell’s law and var. princ.5. Can Snell’s law be derived using the variational principle (or a variational principle “as you

prefer”)? Please explain.

110 Chapt. 16 Variational Principle and Variational Methods

016 qfull 00060 2 5 0 moderate thinking: Schr”od. and var. princ.6. Can the time-independent Schrodinger’s equation be derived using the variational principle?

Please explain.

016 qfull 00070 2 5 0 moderate thinking: convert to matrix eigenproblem7. Convert the braket eigenproblem H |Ψ〉 = E|Ψ〉 to the discrete |uj〉 orthonormal basis

representation by expanding |Ψ〉 in terms of the |uj〉 kets and then operating on the equationwith the bra 〈ui|. Find the matrix representation of the eigenproblem.

016 qfull 00080 1 5 0 easy thinking: solving infinite matrix problem8. Can one literally solve in a numerical procedure an infinite matrix problem: i.e. a problem with

an infinite number of terms to number crunch? Why so or why not?

016 qfull 00090 1 5 0 easy thinking: diagonalization defined9. What is meant by diagonalization in quantum mechanics?

016 qfull 00200 2 5 0 moderate thinking: simple variational methodExtra keywords: simple variational method for excited states

10. The simple variational method can in principle be applied to excited states.

a) Say an unnormalized trial wave function |ψ〉 is orthogonal to all energy eigenstates |φi〉of quantum number less than n, where the eigen-energies increase monotonically withquantum number as usual. Show that Etrial ≥ En where Etrial is the expectation value ofthe Hamiltonian for |ψ〉. When will the equality hold? Remember there is such a thing asdegeneracy.

b) Using the simple variational method for finding excited eigenstate energies isn’t really ofgeneral interest since constructing trial functions with the right orthogonality propertiesis often harder than using the other approaches. However, if the eigenstates have definiteparity, definite parity trial wave functions can be used to determine the lowest eigen-energiesfor wave functions of each kind of parity.

For example, let us consider the simple harmonic oscillator problem in one dimension.We know that the eigenstates are non-degenerate and have definite parity. It is given thatthe ground state has even parity and the first excited state has odd parity. We can usean odd trial wave function and the variational method to approximately determine theenergy of the first excited state. The simple harmonic oscillator eigenproblem in scaleddimensionless variables is

(

− d2

dx2+ x2

)

ψ = Eψ ,

where

x =

√

mω

h−xphy and E =

Ephy

h−ω/2= 2n+ 1 .

The n is the SHO energy quantum number (n runs 0, 1, 2, 3, . . .) and the “phy” stands forphysical. Consider the odd trial wave function

ψ =

x(x2 − c2), |x| ≤ c;0. |x| > c,

where c is a variational parameter. Normalize this trial wave function, evaluate itsexpectation energy, and minimize the expectation energy by varying c. How does thisvariational method energy compare to the exact result which in scaled variables is 3.HINT: There are no wonderful tricks in the integrations: grind them out carefully.

Chapt. 16 Variational Principle and Variational Methods 111

016 qfull 00300 3 5 0 tough thinking: variational hydrogenExtra keywords: (Ha-327:4.1)

11. We know, of course, the ground state for the hydrogenic atom sans perturbations:

ψnℓm =1√4π

(2a−3/2)e−r/a ,

where a = a0/[(m/me)Z] is the radial scale parameter: a0 = h−2/(mee

2) = λCompton/(2πα) =0.529 A is the Bohr radius, m is the reduced mass, and Z is the nuclear charge (Gr-128, 141).But as a tedious illustration of the simple variational method, let us try find an approximateground state wave function and energy starting with the trial Gaussian wave function

ψ = Ae−βr2/a2

.

a) Can we obtain the exact solution with a trial wave function of this form?

temitemb) The varied energy is given by

Ev =〈ψ|H |ψ〉〈ψ|ψ〉 =

∫∞

0[ψ(r)∗Hψ(r)] (4πr2) dr

∫∞

0[ψ(r)∗ψ(r)] (4πr2) dr

,

where H is the Hamiltonian for ℓ = 0 (i.e., the zero angular momentum case) given by

H = − h−2

2m

1

r2∂

∂rr2∂

∂r− Ze2

r.

Note the varied energy form does not require a Lagrange undetermined multiplier since we arebuilding the constraint of normalization into the variation. We, of course, need to evaluate A laterto normalize the minimized wave function. Convert the varied energy expression into a dimensionlessform in terms of the coordinate x = r/a and reduced varied energy ǫv = Ev/[Ze

2/(2a)] =Z−2(m/me)Ev/ERyd ≈ Z−2(m/me)Ev/(13.606 eV). HINT: A further integration transformationcan make the analytic form even simpler.

temitemc) Find the explicit analytic expression for ǫv. Sketch a plot of ǫv as a function of β.HINT: Use an integral table.

d) Now find the minimizing β value and the minimum ǫv. Compare ǫv to exact ground statevalue which is −1 in fact.

016 qfull 01000 3 5 0 tough thinking: non-orthogonal linear variationExtra keywords: method for a two level system.

12. You are given two basis states |1〉 and |2〉 and want to solve a two-dimensional system withHamiltonian H in terms of this basis. The basis is not orthogonal although the basis statesare normalized of course. Recall in this case that the non-orthogonal linear variational methodeigenproblem is

H~c = ES~c ,

where ~c is an unknown eigenvector, E and unknown eigen-energy, and S is the overlap matrix.

Let

H =

(

ε1 VV ε2

)

.

We have assumed that 〈1|H |2〉 = 〈2|H |1〉 and designated these elements by V : i.e., theeigenstates are pure real. This assumption is generality that probably pointless for the cases

112 Chapt. 16 Variational Principle and Variational Methods

where this problem is probably of most interest: i.e., in LCAO method (i.e., linear combinationof atomic orbitals method) for molecular orbitals. We will also assume V < 0 which is alsoappropriate for LCAO, and so avoids needles generality. As a fiducial choice assume ε2 ≥ ε1although all the formulae will not depend this choice in fact. For the overlap matrix let

S =

(

〈1|1〉〈1|2〉〈2|1〉〈2|2〉

)

=

(

1 ss 1

)

.

Your mission Mr. Phelps—if you choose to accept it—is solve for the eigen-energies andeigenvectors. These quantities tend to come out in clumsy forms. So you should try to findnice forms. You may subsume large clumpy expressions into single symbols, but show somerestraint. One trick is to re-origin all the energies: i.e., define

ε =ε1 + ε2

2, −ε′ = ε1 − ε , ε′ = ε2 − ε , V ′ = V − εs , and E′ = E − ε .

Note with our fiducial assumptions ε′ ≥ 0, but all the formula should work for ε′ < 0 too. Notealso that V ′ < 0, V ′ > 0, or V ′ = 0 are all possible now. Now subtract

ε

(

1 ss 1

)

~c

from both sides of the eigenproblem and solve for the primed eigen-energies and the eigen-vectors in terms of the primed quantities. Having found the solutions, you should examine thespecial limiting cases: i.e., ε′ → 0, and s→ 0.

The State Department confesses that it does not know the ideal forms for the solutions andin any case will disavow all knowledge of your activities.

Chapt. 17 Time-Dependent Perturbation Theory


017 qmult 00100 1 1 4 easy memory: Fermi, person identificationExtra keywords: Fermi, person identification

1. Who was Enrico Fermi?

a) An Italian who discovered America in 1492.b) An Italian who did not discover America in 1492.c) An Italian-American biologist.d) An Italian-American physicist.e) Author of Atoms in the Family.

017 qmult 00200 1 4 5 easy deducto-memory: golden ruleExtra keywords: Sc-288

2. “Let’s play Jeopardy! For $100, the answer is: This quantum mechanical time-dependentperturbation result was discovered by Pauli, but named by Fermi.”

a) What is the categorical imperative, Alex?b) What is the sixth commandment, Alex?c) What is the no-fault insurance, Alex?d) What is the iron law, Alex?e) What is the golden rule, Alex?

017 qmult 00300 1 4 5 easy deducto-memory: golden rule validity3. “Let’s play Jeopardy! For $100, the answer is: This aureate time-dependent perturbation result

requires, among other things, that

δElevel separation <<2πh−

t− tchar

<∼∆Ebandwidth ,

where δElevel separation is of order of the separation between energy levels in a continuum bandof energy levels, t− tchar is the time since the perturbation became significant (i.e., tchar), and∆Ebandwidth is the characteristic energy width of the continuum band.”

a) What is 2nd order perturbation, Alex?b) What is 3rd order perturbation, Alex?c) What is the optical theorem, Alex?d) What is Pauli’s exclusion principle, Alex?e) What is Fermi’s golden rule, Alex?

017 qmult 00100 1 1 4 easy memory: exponential decay of state4. Fermi’s golden rule if it applies to transitions to all states from an original state and for all time

after a perturbation is applied (which may be from the time the original state forms) causesthe original state to have:

a) no transitions.

113

114 Chapt. 17 Time-Dependent Perturbation Theory

b) a linear decline in survival probability.c) a power law decline in survival probability.d) an exponential decline in survival probability.e) an instantaneous decline in survival probability.

017 qmult 00800 1 1 5 easy memory: harmonic perturbation, sinusoidalExtra keywords: harmonic perturbation, sinusoidal time dependence

5. Harmonic perturbations have:

a) a linear time dependence.b) a quadratic time dependence.c) an inverse time dependence.d) an exponential time dependence.e) a sinusoidal time dependence.

017 qmult 01200 1 1 1 easy memory: principal value integral6. One can sometimes in integrate over a first order singularity and get a physically reasonable

result. This kind of integral is called:

a) a principal value integral or Cauchy principal value integral.b) an interest integral.c) a capitol integral.d) a bull-market integral.e) a bear-market integral.

017 qmult 02000 1 1 5 easy memory: electric dipole selection rulesExtra keywords: electric dipole selection rules

7. The selection rules for electric dipole transitions are:

a) ∆l = 0 and ∆m = 0.b) ∆l = ±2 and ∆m = ±1.c) ∆l = −1 and ∆m = 1.d) ∆l = ±1 and ∆m = 0.e) ∆l = ±1 and ∆m = 0, ±1.


017 qfull 00010 1 5 0 easy thinking: time-dependent Sch.eqn.1. Is the time-dependent Schrodinger equation needed for time-dependent perturbation theory?

017 qfull 00020 2 5 0 moderate thinking: energy eigenstates2. Are stationary states (i.e., energy eigenstates) needed in time-dependent perturbation theory?

Please explain.

017 qfull 00030 2 5 0 moderate thinking: energy eigenstates3. What is done with the radiation field in quantum electrodynamics.

017 qfull 00100 2 3 0 easy math: Fermi’s golden rule integralExtra keywords: Fermi’s golden rule integral, Simpson’s rule

4. In the simplest version of the derivation of Fermi’s golden rule one uses the integral∫ ∞

−∞

sin2 x

x2dx = π

Chapt. 17 Time-Dependent Perturbation Theory 115

which can be evaluated using complex variable contour integration (Ar-364). One of the featuresof this integral that is used in the justification of the golden rule is that most of the total comesfrom the central bump of the integrand: i.e., the region [−π, π]. It would be good to know whatfraction of the total comes from the central bump. Alas,

Icen =

∫ π

−π

sin2 x

x2dx .

is not analytically solvable.

a) Find an excellent approximate value for Icen. HINTS: It’s probably no good trying tofind a good approximation for the central bump directly since it is most of the total. Anapproximate value could easily turn out to be off by of order |Icen −π|. Try finding a valuefor the non-central bump region.

b) Now—if you dare—evaluate Icen numerically and compare to your analytic result frompart (a). HINT: I use double precision Simpson’s rule myself.

017 qfull 00200 3 3 0 tough math: time dependent perturbation, square wellExtra keywords: (MEL-141:5.3), time dependent perturbation, infinite square well

5. At time t = 0, an electron of charge e is in the n eigenstate of an infinite square well withpotential

V (x) =

0, x ∈ [0, a];∞ x > a.

At that time, a constant electric field E pointed in the positive x direction is suddenly applied.(Note the tildes on charge and electric field are to distinguish these quantities from the naturallog base and energy.) NOTE: The 1-d infinite square-well eigenfunctions and eigen-energiesare, respectively

ψn(x) =

√

2

asin(nπ

ax)

and En =h−2k2

2m=

h−2

2m

(π

a

)2

n2 ,

where n = 1, 2, 3, . . . The sinusoidal eigenfunctions can be expressed as exponentials: letz = πx/a, and then

sin(nz) =einz − e−inz

2i.

a) Use 1st order time-dependent perturbation theory to calculate the transition probabilitiesto all OTHER states m as a function of time. You should evaluate the matrix elementsas explicitly: this is where all the work is naturally.

b) How do the transition probabilities vary with the energy separation between states n andm?

c) Now what is the 1st order probability of staying in the same state n?

017 qfull 00300 3 5 0 tough thinking: usual and general Fermi’s golden rule6. Say we have time-dependent perturbation

H(t) =

0, t < 0;H, t ≥ 0,

and initial state |φj〉, where |φj〉 is the eigenstate belonging to the complete set |φi〉. Thestate at any time t ≥ 0 is |Ψ(t)〉.

116 Chapt. 17 Time-Dependent Perturbation Theory

a) Work out as far as one reasonably can the 1st order perturbation expression for thecoefficient ai(t) in the expansion of |Ψ(t)〉 in terms of the set |φi〉. Include the caseof i = j. HINT: The worked out expression should contain a sine function. Defineωij = (Ei − Ej)/h−.

b) Given i 6= j, find the transition probability (to 1st order of course) from state j to state i.

c) What is this probability at early times when ωijt/2 << 1 for all possible ωij? Describe thebehavior of the probability as a function of time for all times. (You could sketch a plot ofprobability as a function of time.) What is the behavior for ωij = 0 (i.e., for transitions todegenerate states)?

d) You want to calculate the summed probability of transition to some set of states (whichmay not be all possible states) that are dense enough in energy space to form a quasi-continuum or even a real continuum of states. The set does not include the initial state j.The summed probability for energy interval Ea to Eb can be approximated by an integral:

P (t) =∑

i6=j

Pi(t) ≈∫ Eb

Ea

P (E, t)ρ(E) dE ,

where ρ(E) is the density of states per unit energy and where the time-independent partof the matrix element Hij is replaced by H(E) which is a continuous function of energy.

What is the total transition probability to all states in the set assuming integrand isonly significant in small region near Ej . The region is small eneough that |H(E)|2 andρ(E) can be taken as constants and that the limits of integration can be set to ±∞. (Noteyou will probably need to look up a standard definite integral.)

In fact, 90 % of the integral (assuming |H(E)|2 and ρ(E) constant) comes from theenergy range [Ej − 2πh−/t, Ej + 2πh−/t]. (Can you show this by a numerical integration?No extra credit for doing this: insight is the only reward.) We can see that at sometime the 90 %-range will be so narrow that the approximation |H(E)|2 and ρ(E) constantwill probably become valid. They should clearly be evaluated at Ej . Practically, thisoften means that the approximation becomes valid when almost all of the transitions areto nearly degenerate states. Of course, the 90 %-range can become so narrow that theapproximation of a continuum of states breaks down and then the integration becomesinvalid again. In fact, for the integration to be valid we require |H(E)|2 and ρ(E) to beconstant over ∆E such that ∆E >∼ 2πh−/t and that the energy separation δE between the

final satisfy δE << 2πh−/t. Thus we require

δE << 2πh−/t <∼∆E .

What is the rate of transition for (i.e., time derivative of) the total transitionprobability? The transition rate result is one of the usual forms of Fermi’s golden rule.Although it is restricted in many ways, it is still a very useful result: hence golden.

e) Let’s see if we can derive a generalized golden rule without the restriction that theperturbation is constant after a sudden turn-on. To do this assume that the perturbationHamiltonian has the form

H(t) = Hf(t) ,

where H is now constant with time and f(t) is a real turn-on function with the propertythat f(t) is significant only for t ≥ tch, where tch is a characteristic time for turn-on. Lettime zero be formally set to −∞ for generality.

First, derive Pi(t) with explicit integrals. Second, assume again that there is a continuumor quazi-continuum of states (of which state i is one) with density of states ρ(E) and that Hij

Chapt. 17 Time-Dependent Perturbation Theory 117

can be replaced by H(E). Third, argue that the time integrals must be sharply peaked functionsof E about the initial E = Ej for t ≥ tch. Fourth, re-arrange the integrals and integrate overenergy making use of the third point. You can then make use of the result

δ(x) =

∫ ∞

−∞

e±ikx

2πdk ,

where δ(x) is the Dirac delta function (Ar-679). What is the total transition probability fort ≥ tch? What is the total transition rate t ≥ tch? When does this generalized golden rulereduce to Fermi’s golden rule?

Chapt. 18 The Hydrogenic Atom and Spin


018 qmult 00100 1 1 4 easy memory: spin-orbit interaction, hydrogenic atom

Extra keywords: spin-orbit interaction, hydrogenic atom

1. What is the main internal perturbation/perturbations preventing the spinless hydrogeniceigenstates from being the actual ones?

a) The Stark effect. b) The Zeeman effect. c) The Stern-Gerlach effect.d) The spin-orbit interaction and the relativistic perturbation.e) The Goldhaber interaction.

018 qmult 00200 2 4 5 moderate deducto-memory: orbital ang. mom., spin

Extra keywords: orbital angular momentum, spin, total angular momentum

2. The scalar product of operators ~L · ~S equals

a) J2.

b) (~L+ ~S) · (~L + ~S).

c) (~L− ~S) · (~L − ~S).

d) (J2 + L2 + S2)/2.

e) (J2 − L2 − S2)/2.

018 qmult 00300 1 4 3 easy deducto-memory: spin-orbit good quantum numbers

Extra keywords: spin-orbit interaction, good quantum numbers

3. “Let’s play Jeopardy! For $100, the answer is: The spin-orbit interaction causes the eigenstatesof the real hydrogen atom to be mixtures of the Ψnℓm states, but one Ψnℓm state is usuallyoverwhelmingly dominant.

a) Why are the quantum numbers n, ℓ, and m perfectly rotten, Alex?

b) Why are the quantum numbers n, ℓ, and m only approximately rotten, Alex?

c) Why are the quantum numbers n, ℓ, and m only approximately good, Alex?

d) Why are the quantum numbers n, ℓ, and m only indifferent, Alex?

d) Why are the quantum numbers n, ℓ, and m dependent on a recount in Palm Beach, Alex?


018 qfull 00500 1 3 0 easy math: fine-structure energy levels

1. The hydrogen atom energy level energies corrected for the fine structure perturbations (i.e., therelativistic and spin-orbit perturbations) is

E(n, ℓ,±1/2, j) = −ERyd

n2

m

me

[

1 +α2

n2

(

n

j + 1/2− 3

4

)]

,

118

Chapt. 18 The Hydrogenic Atom and Spin 119

where n is the principal quantum number, ℓ is the orbital angular momentum quantum number,±1/2 is allowed variations of j from ℓ, j (the total angular momentum quantum number) is aredundant parameter since j = max(ℓ± 1/2, 1/2) (but it is a convenient one),

ERyd =1

2mec

2α2

is the Rydberg energy, me is the electron mass, α ≈ 1/137 is the fine structure constant, and

m =memp

me +mp

is the reduced mass with mp being the proton mass. The bracketed perturbation correctionterm is

α2

n2

(

n

j + 1/2− 3

4

)

which is of order α2 ≈ 10−4 times smaller than the unperturbed energy. Show that theperturbation term is always negative and reduces the energy from the unperturbed energy:i.e., show that

n

j + 1/2− 3

4> 0

in all cases.

Chapt. 19 Symmetrization Principle


019 qmult 00100 1 4 5 easy deducto-memory: symmetrization principle1. “Let’s play Jeopardy! For $100, the answer is: It is the quantum mechanics POSTULATE that

wave functions for identical particles must be symmetrized: i.e., symmetric for bosons (integerspin particles) and antisymmetric for fermions (half-integer spin particles) under the exchangeof any two identical elementary particles. The postulate evolved in the 1920s from the workof Pauli, Fierz, Weisskopf, Heisenberg, Dirac, and others: there seems to be no one discoverer.An immediate corollary of the postulate is that composite particles obey it too even if they arenot identical (because they are in different states and/or subject to different perturbations) aslong as their constituent elementary particles are identical.

Actually one needs to define exchange. For simplicity, we will only consider elementaryparticles and neglect spin coordinates—which take a lot of explaining. Particles are all givenlabeled coordinates. To “exchange” particles is to exchange their labeled coordinates in wavefunction slots. A slot in notation is just where one puts a function argument in functionexpression. Mathematically a slot is how the function depends on the argument value put inthe notational slot. For example, wave function

ψ(x1, x2)

has two slots: the first has coordinate x1 for particle 1 in it and the second coordinate x2 forparticle 2 in it. The wave function in general has a different functional dependence on thearguments in the two slots. So to “exchange” particles is NOT really about considering thesystem with two particles in exchanged in particular positions in space, but is about consideringthe wave function with an exchanged functional dependence on the particles.

Now if particle 1 and particle 2 were distinct particles, then the wave function ψ(x2, x1) isnot a physically real wave function in general (although it may be in special cases) since it saysthat the particles behave exactly alike even though they are distinct—there’s a contradiction:how can particles be distinct and behave exactly alike in all cases?

On the other hand, ψ(x2, x1) is allowed for identical particles in general since they areidentical. But the under-discussion POSTULATE gives the restriction

ψ(x2, x1) = ±ψ(x1, x2) ,

where the upper case is for bosons and lower case for fermions. Arguments that the postulatemust be true for identical particles because they are identical are specious. If the postulatefollows from arguments, then it is not a postulate. Instead one has to say that a feature ofidentical particles in quantum mechanics is that they obey the postulate.”

What is , Alex?

a) Born’s hypothesis b) Schrodinger’s dilemma c) Dirac’s paradox d) Wigner’s last stande) the symmetrization principle or postulate

019 qmult 00110 1 1 3 easy memory: degeneracy and symmetrization principle

120

Chapt. 19 Symmetrization Principle 121

2. As strange as the symmetrization principle seems, the world would be even stranger without itsince then we would have among other things which among leaves us wonderingwhat to do with those states and would make statistical mechanics as it is now formulatedimpossible.

a) no degeneracy of stationary states at allb) a finite degeneracy of stationary states for many systemsc) an uncountable infinite degeneracy of stationay states for many systemsd) no stationary states at all e) no energy at all

019 qmult 01000 1 4 5 easy deducto-memory: Bose-Einstein condensateExtra keywords: References Gr-216, CDL-1399, Pa-179

3. “Let’s play Jeopardy! For $100, the answer is: The name for the state of a system of all identicalbosons when the bosons all settle into the ground state.”

a) What is a Hermitian conjugate, Alex?b) What is a Hermitian condensate, Alex?c) What is a Rabi-Schwinger-Baym-Sutherland-Jeffery degeneracy, Alex?d) What is just another state, Alex?e) What is a Bose-Einstein condensate, Alex?


019 qfull 00100 2 5 0 moderate thinking: permutation operator1. The permutation operator P has the seemingly arbitrary, but well defined, property that

Pf(x1, x2) = f(x2, x1) ,

where f(x1, x2) is a general complex function of two pure real arguments.

a) Show that the permutation operation and the complex conjugation operation commute:i.e., show that

[Pf(x1, x2)]∗ = P [f(x1, x2)

∗] .

HINT: Decompose f(x1, x2) into real and imaginary parts.

b) Show from the definition of the Hermitian conjugate,

〈φ|Q|ψ〉 = 〈ψ|Q†|φ〉∗

(where Q is any operator), that P is a Hermitian operator: i.e., that P = P †. HINT:Recall that for two spatial dimensions

〈φ|Q|ψ〉 =

∫

1

∫

2

φ(x1, x2)∗Qψ(x1, x2) dx1 dx2 .

c) Solve for ALL the eigenvalues of P .

d) Show that any function f(x1, x2) can be expanded in eigenfunctions of P , and thus theeigenfunctions of P form a complete set for the space of functions of two argumentsincluding wave function spaces of two arguments. Since P is Hermitian and has a completeset of eigenfunctions for any wave function space of two arguments, it is formally a quantummechanical observable.

122 Chapt. 19 Symmetrization Principle

e) Given that A(x1, x2) is an operator, show that

PA(x1, x2)f(x1, x2) = A(x2, x1)Pf(x1, x2) .

Recall that operators act on everything to the right—except, of course, when they don’t:but that situation is usually (but not always) made explicit with brackets. Do P and Acommute in general? When do they commute?

f) Show that P and the Hamiltonian for identical particles,

H = − h−2

2m

∂2

∂x21

− h−2

2m

∂2

∂x22

+ V (x1, x2) ,

commute. Show that if ψ(x1, x2) is an eigenstate of the Hamiltonian, then Pψ(x1, x2) isan eigenstate. If ψ(x1, x2) is non-degenerate in energy, is Pψ(x1, x2) a physically distinctstate? Show that there are only two possibilities for what Pψ(x1, x2) is?

g) Given that P andH commute, show that P is a constant of the motion as far as Schrodingerequation evolution goes.

019 qfull 00200 2 5 0 moderate thinking: exchange degeneracy2. Say you have two distinct, ORTHONORMAL single-particle energy eigenstates ψa(x) andψb(x) and you wish to construct from them a two-particle energy eigenstate for two identicalspinless particles: particles 1 and 2. One possibility is

ψ(x1, x2) = ψa(x1)ψb(x2) .

A more general two-particle state is

ψ(x1, x2) = c1ψa(x1)ψb(x2) + c2ψa(x2)ψb(x1) .

NOTE: We are only discussing spatial eigenstates here, not full time dependent wave functions.

a) Find the condition on the coefficients c1 and c2 such that ψ(x1, x2) is normalized.

b) How many energy degenerate states can be formed by the various choices of c1 and c2consistent with normalization? This degeneracy is called the exchange degeneracy (CDL-1376). Would this degeneracy exist if the particles were distinct? Why or why not?

c) All of statistical mechanics and atomic spectroscopy (where transition rates depend ondegeneracies) tell us that the vast degeneracy found in the part (b) answer does not exist innature. No limitation on this degeneracy can be derived. But one can postulate a limitation.Given that the particles are identical, one natural way to mostly kill the degeneracy is topostulate that the only allowed choices of c1 and c2 are those that treat terms in theexpression for ψ(x1, x2) equally. Given this postulate, find all the allowed c1 and c2 pairsthat are distinct to within a global phase factor. (Note if you multiplied by a c1 and c2 pairby eiζ you would not have created a physically distinct pair.) HINT: Remember that aset of wave functions that differ only by a global phase factor are actually only one physicalwave function.

d) Evaluate 〈x1〉 and 〈x2〉 for the allowed c1 and c2 values found in the part (c) answer. Youcan take as given

〈x〉a = 〈ψa|x|ψa〉 and 〈x〉b = 〈ψb|x|ψb〉 .

Formally the operators x1 and x2 are quantum mechanical observables. But would theexpectation values 〈x1〉 and 〈x2〉 be INDIVIDUALLY observable in fact if for c1 and c2values other than those allowed by the part (c) answer?


e) Show that the wave functions with the allowed coefficients are eigenfunctions of thepermutation operator P which has the effect on the wave function that

Pψ(x1, x2) = ψ(x2, x1) .

019 qfull 02000 2 5 0 moderate thinking: symmetrizationExtra keywords: symmetrization of orthonormal single-particle states.

3. Say |ai〉 and |bi〉 are ORTHONORMAL single-particle states, where i is a particle label. Thelabel can be thought of as labeling the coordinates to be integrated or summed over in an innerproduct: see below. The symbolic combination of such states for two particles, one in a andone in b is

|12〉 = |a1〉|b2〉 ,where 1 and 2 are particle labels. This combination is actually a tensor product, but let’s notworry about that now. The inner product of such a combined state is written

〈12|12〉 = 〈a1|a1〉〈b2|b2〉 .

If one expanded the inner product in the position and spinor representation assuming the wavefunction and spinor parts can be separated (which in general is not the case),

〈12|12〉 =

[∫

ψa(x1)∗ψa(x1) dx1 ( c∗1+ c∗1− )a

(

c1+c1−

)

a

]

×[∫

ψb(x2)∗ψb(x2) dx2 ( c∗2+ c∗2− )b

(

c2+c2−

)

b

]

.

A lot of conventions go into the last expression: don’t worry too much about them.

a) Let particles 1 and 2 be NON-identical particles. What are the two simplest and most obviousnormalized 2-particle states that can be constructed from states a and b? What happens ifa = b (i.e., the two single-particle states are only one state actually)?

b) Say particles 1 and 2 are identical bosons or fermions. What is the simplest and most obviousnormalized 2-particle state that can be constructed in either case allowing for the possibilitythat a = b (i.e., the two single-particle states are only one state actually)? What happens ifa = b, for fermions.

019 qfull 02100 1 5 0 easy thinking: triplet singletExtra keywords: (Gr-181:5.3)

4. Say that we have obtained four orthonormal single particle eigenstates:

ψa(~r)χ+

ψa(~r)χ−

ψb(~r)χ+

and

ψb(~r)χ−

where the spinors are

χ+ =

(

10

)

or χ− =

(

01

)

.

To label a state i’s coordinates by a descriptive label one can write for example

ψa(i)(~r)χ+(i) .


Construct ????

019 qfull 02200 3 5 0 tough thinking: 2-particle infinite square wellExtra keywords: (Gr-182:5.4)

5. The set of individual eigen states for a 1-dimensional, infinite square well confined to [0, a] canbe written |n〉 where n = 1, 2, 3, . . . The energies of the states are given by

E(n) =h−2

2m

(π

a

)2

n2

(e.g., Gr-26). For convenience Ered(n) = n2 can be called the reduced energy of state n.a) Say we have two non-interacting particles a and b in the well. Write write down the

Hamiltonian for this case. The particles have the same mass m, but are not necessarilyidentical.

b) The reduced energy of a 2-particle state that satisfy the Schrodinger equation of part (a)can be written

Ered(n1, n2) = n21 + n2

2 .

Write a small computer code to exhaustively calculate the possible reduced energy levelsup to and including Ered = 50 and the n1 and n2 combinations that yield these energies.The code should also calculate the degeneracy of each energy for the cases of non-identicalparticles, bosons, and fermions. I’ll left you off easily, accidental degeneracies can beidendified by eye. (Note: An accidental degeneracy is when a distinct pair of n values (i.e.,a pair not counting order) gives the same reduced energy.)

c) Write down the normalized vector expressions for all the 2-particle states up to the 4thallowed energy level for the cases of non-identical particles, identical bosons, and identicalfermions. Just to get you started the non-identical particle ground state is

|a1, b1〉 = |a1〉|b1〉 with Ered = 2 .

019 qfull 02300 3 5 0 tough thinking: exchange forceExtra keywords: (Gr-182)

6. The exchange force is a pseudo-force that arise because of the symmetry postulate of quantummechanics. Say we have orthonormal individual particle states |a〉 and |b〉. If we havedistinguishable particles 1 and 2 in |a〉 and |b〉, respectively, the net state is

|1, 2〉 = |a1〉|b2〉 .

Of course, each of particles 1 and 2 could be in linear combinations of the two states. In thatcase the combined state would be a four term state. But we have no interest in pursuing thatdigression at the moment. Now 2 indistinguishable particles in states |a〉 and |b〉 have no choice,but to be in a combined symmetrized state by the symmetry postulate:

|1, 2〉 =1√2

(|a1〉|b2〉 ± |b1〉|a2〉) ,

where the upper case is for identical bosons and the lower case for identical fermions. If the twostates are actually the same state |a〉, then the state for distinguishable particles and bosons isthe same

|1, 2〉 = |a1〉|a2〉

and no state is possible for fermions by the Pauli exclusion principle.


Note products of kets are actually tensor products (CDL-154). In taking scalar products,the bras with index i (e.g., 1 or 2 above) act on the kets of index i. For example, for the state|1, 2〉 = |a1〉|a2〉 the norm squared is

〈1, 2|1, 2〉 = 〈a1|a1〉〈a2|a2〉 .

The fact that identical particles must be combined symmetrized states means that theirwave functions will be more or less clumped depending on whether they are bosons or fermionsthan if they could be fitted into simple product states like distinguishable particles. (Note weare not bothering with the complication of spin for the moment. One could say we are lettingall the spins point up for example). This clumping/declumping effect is called the exchangeforce. Obviously, it is not really a force, but rather a result of the requirements on allowedstates. Still for some practical purposes one can certainly consider it as force.

a) Expand 〈∆x2〉 = 〈(x1 − x2)2〉.

b) For the given states, determine 〈∆x2〉 for distinguishable particles and, for the case of beingonly one single-particle state |a〉, for indistinguishable bosons.

c) For the given states constructed from distinct single-particle states, determine 〈∆x2〉 forindistinguishable bosons and fermions.

019 qfull 02400 2 5 0 moderate thinking: exchange force and infsq wellExtra keywords: (Gr-185:5.5) and the infinite square well

7. Imagine two non-interacting particles in an infinite square in the range [0, a]. Recall the eigen-functions for this case are

ψn =

√

2

asin(nπ

ax)

for n = 1, 2, 3, . . .. Recall also the results of the Gr-182 and Gr-29:2.5 questions.

a) Say the particles are distinguishable and are in states n andm. What is 〈∆x2〉 = 〈(x1−x2)2〉

for this case? What is it if n = m?

b) Say the particles are identical bosons/fermions and are in the only allowed combination ofstates n and m. What is 〈∆x2〉 = 〈(x1 − x2)

2〉 for this case? What is it if n = m?

019 qfull 02500 2 3 0 mod math: coupled harmonic oscillatorExtra keywords: two identical particles, exact solution

8. There are two particles subject to separate simple harmonic oscillator (SHO) potentials. Theyare also coupled by a SHO interaction. The full Hamiltonian is:

H =p21

2m1+

p22

2m2+

1

2m1ω

2x21 +

1

2m2ω

2x22 +

1

2k(x1 − x2)

2 ,

where k > 0 which in this context means the interaction is attractive.

a) Transform to the center-of-mass-relative (CM-REL) coordinates (showing all the steps)and show that the Hamilton separates into a center-of-mass (CM) SHO Hamiltonian and arelative (REL) SHO Hamiltonian. Does the problem have an exact solution? Write downthe general expression for the eigen-energies of the total stationary states in terms of theSHO quantum numbers nCM and nREL for the respective CM and REL parts. Define ω asthe angular frequency of the REL energies.

b) Next write the expression for the eigen-energies in the case that k = 0. Define a newquantum number n that alone gives the eigen-energy and the degeneracy of the eigen-energy. What is the degeneracy of an eigen-energy of quantum number n.


c) Now assume that k > 0, but that k/(µω2) << 1. Write down a first order correct expressionfor the energy in terms of n and nREL. Give a schematic energy-level diagram.

d) Now assume that k/(µω2) >> 1. Give a schematic energy-level diagram in this case.

e) Now assume that the two particles are identical spin-0 bosons. Note that identical meansthey now have the same mass. Given the symmetry requirement for boson states, whichsolutions (specified by the nCM and nREL quantum numbers) are not physically allowed?

f) Now assume that the two particles are identical spin-1/2 fermions. Note again that identicalmeans they now have the same mass. But also note they arn’t electrons. Their interactionsare determined by the given Hamiltonian only. Because the particles are spin-1/2 fermions,the eigen wave functions for system must be multiplied by appropriate eigen-spinors tospecify the full eigenstate. Given the antisymmetry requirement for fermion states, whatrestrictions are put on the wave function and spinor quantum numbers of an eigenstate?

019 qfull 02600 1 5 0 easy thinking: symmetrization, Slater determinant

Extra keywords: (Gr-187:5.7)

9. Say that you solve a Schrodinger equation for N identical particles to get the normalized wavefunction ψ(~r1, ~r2, ~r3, . . . , ~r N ). How would you symmetrize the wave function for bosons? Thenhow would you symmetrize for fermions all in the spin-up state so that you don’t have spinorsto complicate the question? How would you normalize the wave function?

019 qfull 02700 1 5 0 easy thinking: doubly excite He decay

Extra keywords: (Gr188.58a)

10. Say you put two electrons into the n = 2 principle quantum number shell of a neutral heliumatom and immediately one electron is ejected and the other decays to the ground of the He+ ion.What approximately is the kinetic energy of the ejected electron. NOTE: Without a detailedspecification of the doubly-excited helium atom we cannot know exactly what the energies of theexcited electrons are. There are two simple approximate choices for their energies: 1) assumethat the energy levels of the singly-excited helium atom apply (see, e.g., Gr-189); 2) assume thatthe Z = 2 hydrogenic energy levels apply. The first choice is probably most in error because itassumes too much electron-electron interaction: the electrons may further apart in the actualdoubly-excited state; but, in fact, where they are depends on exactly what doubly excited statethey are in. The 2nd choice is certainly wrong by assuming zero electron-electron interaction.

019 qfull 02900 2 5 0 moderate thinking: helium with bosons

Extra keywords: (Gr-188:5.9)

11. Describe qualitatively how the helium atom energy level diagram would plausibly change underthe following conditions.

a) Say the electrons were spin zero bosons.

b) Say the electrons were spin 1/2 bosons—a contradiction in postulates, but for the sake ofargument have it so.

c) Say the electrons were spin 1/2 fermions, but were quantum mechanically distinguishableparticles. HINT: In this case the answer is going to be pretty much indefinite.

019 qfull 03000 2 5 0 moderate thinking: Bose-Einstein counting

Extra keywords: See Po-13 and Po-47

12. In statistical mechanics, the symmetrization requirement on identical bosons enters in the waythat probabilities are assigned to the global states they can form. We will investigate howsymmetrization manifests itself in this case.


a) Say you had g single-particle states and n distinct particles. How many distinct globalstates can you form? What is the probability of each global state assuming that each hasequal probability?

b) Now a trickier case. Say you had g single-particle states and n identical particles. Theprobability pi that a particle goes into single-particle state i is INDEPENDENT of whatthe other particles do: note

∑gi=1 pi = 1, of course. You can construct all possible global

states by inserting one particle at a time into the system—can you imagine a global statethat cannot be so constructed? Say you do insert the n particles one at a time to the system.The probability of an n-particle global state formed by the insertion sequence ijk . . . ℓ ispipjpk . . . pℓ which has n factors, of course. But because the particles are identical, each(distinct) global state can be constructed in general by multiple insertion sequences. Howmany distinct insertion sequences for n particles correspond to a single global state withoccupation number set ni? If all the pi are equal, what is the probability of a globalstate with occupation number set ni formed by random insertion of particles?

The sum of the probabilities for all insertion sequences is 1. Why must this be soon general grounds? Now prove more explicitly that the sum of all inserttion sequenceprobabilities is 1. HINT: Consider

1 =

(

g∑

i=1

pi

)n

and a proof by induction.

c) Now in the part (b) answer, we didn’t find out how many distinct global states there were.To find this out you need a different counting procedure. Let’s consider finding all possibleglobal states given the following conditions. Imagine that all n particles were distinct andthat the order in which you choose the single-particle states to slot them into also matters.To start with you must select a state: you can’t put a particle in a non-state. Then proceedselecting a particle for the current state or a new state until you are out of particles andstates. Now did the order of the states matter or the order of the choice of particles?

d) Now for classical, non-interacting particles randomly slotted into single-particle states,the probability of each global state is as determined in part (b). Quantum mechanicalnon-interacting bosons do not act like classical particles. Because of the symmetrizationprinciple—in a way the instructor has never found out—each distinct global state has equalprobability. What is this probability for n bosons in g single-particle states? Say that wehave all n bosons in one single-particle state. What is the classical probability of this globalstate? Which is larger the classical probability or the boson probability? What does thelast result suggest about the random distributions of bosons relative to classical randomdistributions?

e) Consider two identical coins—say quarters. How many distinct global physical states canbe made given that the single-coin states are head and tail? Now toss them up togetherin a completely randomizing way 36 times. Count the number of distinct global states ofeach kind that you get? Do the probabilities of each distinct global state appear to beclassically random or boson random?

Chapt. 20 Atoms


020 qmult 00100 1 1 1 easy memory: atom defined1. An atom is a stable bound system of electrons and:

a) a single nucleus.b) two nuclei.c) three nuclei.d) a single quark.e) two quarks.

020 qmult 01000 1 4 1 easy deducto-memory: central potential2. “Let’s play Jeopardy! For $100, the answer is: The overwhelmingly favored way to solve for the

electronic structure of atoms in quantum mechanics.”

a) What is the central potential approximation, Alex?b) What is the non-central potential approximation, Alex?c) What is the grand central approximation, Alex?d) What is the atom-approximated-as-molecule method, Alex?e) What is the electrons-as-bosons approximation, Alex?


020 qfull 00100 1 5 0 easy thinking: spectrum of He IIExtra keywords: (Gr-188:58b)

1. Describe the spectrum of He II (i.e., singly-ionized helium or He+) sans perturbations..

020 qfull 00200 3 3 0 tough math: helium atom 1st order perturbationExtra keywords: (Gr-188:5.10)

2. If one neglects the electron-electron interaction of the helium atom then the spatial ground stateis just the product of two hydrogenic states:

ψ(~r1, ~r 2) = ψ100(~r 1)ψ100(~r 2) =1

πa3He/8

e−2(r1+r2)/a =8

πa3e−2(r1+r2)/a ,

where aHe = a/Z = a/2 is the helium Bohr radius and a is the standard Bohr radius (see, e.g.,Gr-137–138 and Gr-187). The 1st order perturbation correction to the helium atom groundstate is given by

〈H ′〉 ,where H ′ is the perturbation Hamiltonian: i.e.,

e2

4πǫ0

1

|~r1 − ~r2|

128

Chapt. 20 Atoms 129

in MKS units (see, e.g., Gr-187) ore2

|~r1 − ~r2|in Gaussian CGS units. Note that we use e for both fundamental charge unit and the exponentialfactor: this is conventional of course: context must decide which is which.

a) Analytically calculate⟨

1

|~r1 − ~r2|

⟩

.

HINTS: Set

|~r1 − ~r2| =√

r21 + r22 − 2r1r2µ ,

where µ = cos θ is the angle cosine between the vectors. Integrate over all ~r2 space firsttaking ~r1 as the z axis for spherical coordinates. It helps to switch to dimensionalessvariables earlier on. There are no specially difficulties or tricks: just a moderate numberof steps that have to be done with tedious care.

b) Now the expression for a hydrogenic energy level, sans perturbations, is

En = −1

2mec

2α2Z2

n2= −Z

2Eryd

n2≈ −13.6Z2

n2eV ,

where me is the electron mass, α is the fine structure constant, Z is the nuclear charge,and Eryd ≈ 13.6 eV is the Rydberg energy (see, e.g., Ga-197). In Gaussian CGS units

α =e2

h−cand a =

h−mecα

(e.g., Ga-199). What is the energy of the helium atom ground state in terms of Rydbergenergies and eVs?

020 qfull 01000 2 5 0 moderate thinking: quantum defectsExtra keywords: (MEL-220:9.1) Needs some more work, particularly (b)

3. Excited states of atoms can usually be approximated as merely promoting a valence electron toa single-particle state at a higher energy than any of the ones used in the ground state. For highenergy single-particle states one often finds that their energies form a quasi-Rydberg series: i.e.,

Enℓ ≈ − ERyd

(n− µnℓ)2,

where n is the principal quantum number, ℓ is the angular momentum quantum number, ERyd =13.606 eV is the Rydberg energy, and µnℓ is the quantum defect. I suppose quantum defect getsits name since it accounts for a “defect” in the quantum number. When quantum defects aresmall the wave functions will be quasi-hydrogen-like and hydrogen-like approximations can beused with some confidence—which is often an immense simplification. (Note I use hydrogen-like,not hydrogenic: hydrogen-like implies that the central potential is like e2/r, whereas hydrogenicimplies the central potential is like ZNe

2/r—at least that’s the way it is in this question. Varioususes can be made of the quantum defect parameterization of energy (e.g., Mi-97).

In understanding quantum defects, three facts are useful to know. First, the single-particlepotential well outside of the core approximates

V (r) = −e2

r

130 Chapt. 20 Atoms

for a neutral atom. Second, near the nucleus of the atom, the single-particle potentialapproximates the bare nucleus potential

V (r) = −ZNe2

r,

where ZN is the nuclear charge. Third, wave functions in central potentials for small radius tendto go as rℓ. This result is rather robust since it is true of hydrogenic wave functions (MEL-57;Ar-622) and spherical Bessel functions (Ha-37) which are the radial solutions of the infinitespherical well. Only s states (i.e., ℓ = 0 states) have non-zero probability amplitude at r = 0.

a) Find the formula for the quantum defect in terms of the energy of the energy level. Thencompute the quantum defects for sodium and lithium given the following table. HINT: Acomputer program would lessen the labor.

Table: Observed Energy Levels in Electronvolts (not to modern accuracy)

n s p d f

Li

2 −5.3923 −3.5442 . . . . . .3 −2.0188 −1.5577 −1.5133 . . .4 −1.0509 −0.8701 −0.8511 −0.85025 −0.6432 −0.5544 −0.5446 −0.5434

Na

3 −5.1397 −3.0359 −1.5223 . . .4 −1.9480 −1.3863 −0.8557 −0.85075 −1.0229 −0.7946 −0.5472 −0.54456 −0.6297 −0.5150 −0.3797 −0.3772

b) As you can see from your table of quantum defects, the quantum defects are NOT nearlyzero in general as they would be if the states were almost exactly hydrogen-like at theirprincipal quantum number. Also the quantum defects are NOT just equal to the principalquantum number of highest core shell ncore: i.e., they are not 1 for Li and 2 for Na. Thus,the quantum defect shows that the outer states are not hydrogen-like either for their actualprincipal quantum number n nor at an effective principal quantum number n−ncore. Whyare these two a priori guesses for quantum defects wrong even though the potential is closeto that of hydrogen with ZN = 1 in the region where the bulk of the probability for outerstates is located.

c) Give a reason why quantum defects should have positive values if they are significantlylarge. Why might small quantum defects be negative?

d) Why do quantum defects decrease with ℓ?

e) Explain why the Na quantum defects tend to be larger than the Li quantum defects?

f) Give a reason why quantum defects may not vanish as n→ ∞.

Chapt. 21 Molecules


021 qmult 00100 1 1 1 easy memory: molecule defined1. A stable bound system of electrons and more than one nucleus with some specific recipe for the

numbers of each kind of nuclei is:

a) a molecule.b) an atom.c) a nucleon.d) a fullerene.e) a baryon.

021 qmult 00100 1 4 1 easy deducto-memory: atoms bound in molecules2. “Let’s play Jeopardy! For $100, the answer is: Because they are formed by bonding atoms and

dissociate into atoms?”

a) What is one good reason for thinking of molecules as bound atoms, Alex?b) What are three bad reasons for thinking of molecules as bound atoms, Alex?c) What is no reason at all for thinking of molecules as bound atoms?d) What is ambiguous answer, Alex?e) What is am-Piguous answer, Alex?

021 qmult 00300 2 1 2 easy memory: molecular energy scales3. Given electron mass m and typical nuclei mass M , the ratio of electronic, vibrational, and

rotational energies for a molecule is of order:

a) 1 : 1 : 1.b) 1 : (m/M)1/2 : (m/M).c) 1 : (m/M) : (m/M).d) 1 : (m/M)1/4 : (m/M)1/2.e) 1 : (m/M)1/4 : (m/M)1/3.

021 qmult 00400 1 4 4 easy deducto-memory: Born-Oppenheimer approx.Extra keywords: (Ba-473)

4. “Let’s play Jeopardy! For $100, the answer is: This approximation allows one to treat thenuclei in atoms as though they interacted only with an effective potential constructed fromactual potentials and the electronic kinetic energy (i.e., the total electronic energy).”

a) What is the Alpher-Behte-Gamow recipe, Alex?b) What is the Einstein-Woody-Allen approximation, Alex?c) What is the linear combination of atomic orbitals method, Alex?d) What is the Born-Oppenheimer approximation, Alex?e) What is the tight-binding approximation, Alex?

021 qmult 00500 1 4 5 easy deducto-memory: tight-binding theory

131

132 Chapt. 21 Molecules

5. “Let’s play Jeopardy! For $100, the answer is: It posits that overlapping wave functions ofbound atoms can be treated to some degree in terms of the orbitals of isolated atoms.”

a) What is the genetic algorithm method, Alex?

b) What is the linear variational method, Alex?

c) What is tight-binder theory, Alex?

d) What is tight-wadding theory, Alex?

e) What is tight-binding theory, Alex?

021 qmult 00600 1 1 3 easy memory: LCAO

6. In LCAO (linear combinations of atomic orbitals method) one uses atomic orbitals as a non-orthonormalized basis set for constructing:

a) bound states of nucleons.

b) bound states of photons.

c) inter-atomic bonding and anti-bonding states.

d) intra-atomic stationary states.

e) property-law violating beach-front homes in California.


021 qfull 00100 2 5 0 moderate thinking: universal sp-coupling parameters

Extra keywords: See Ha-95

1. Harrison (Ha-95) presents “universal” sp-bond matrix elements or coupling pararameters:

〈s1|H |s2〉 = Vssσ = −π2

8

h−2

md2,

〈pz1|H |pz2〉 = Vppσ =3π2

8

h−2

md2,

〈s1|H |pz2〉 = Vspσ =π

2

h−2

md2,

and

〈px1|H |px2〉 = 〈py1|H |py2〉 = Vppπ = −π2

8

h−2

md2= Vssσ ,

where 1 and 2 denote two atoms aligned along the z-axis, H is the single-particle Hamiltonianowing to the cores of the two atoms, the |s1〉, etc., are single-particle atomic orbitals (radialparts of some sort times the cubical harmonics for the angular parts) oriented relative to acommon set of axes, d is the inter-nuclei separation, and Vspσ has a π, not π2. The s and psubscripts on the V ’s indicate the atomic orbitals in the coupling and the σ and π indicate theindicates the quantum numbers m2 of L2

z operator of the molecular orbitals that result fromthe coupling of the different states: σ is for m2 = 0 and π is for m2 = 1.

The reason for the complication of using the eigenvalues of the L2z operator rather than the Lz

operator is that the |px〉 and |py〉 cubical harmonics are eigenstates of L2z, but not of Lz. Recall

the lowest quantum number spherical harmonics Yℓ,m are

Y0,0 =1√4π

, Y1,0 =

√

3

4πcos(θ) , and Y1,±1 =

√

3

8πsin(θ)e±imφ ,

Chapt. 21 Molecules 133

where ℓ is the L2 quantum number, m is the Lz quantum number, θ is the angle from the z-axis,and φ is the azimuthal angle. The cubical harmonics are defined by

|s〉 = Y0,0 =1√4π

,

|px〉 =Y1,1 + Y1,−1√

2=

√

3

4π

x

r,

|py〉 =Y1,1 − Y1,−1

i√

2=

√

3

4π

y

r,

and

|pz〉 = Y1,0 =

√

3

4π

z

r.

a) Verify that polar plots of the “p” cubical harmonics are a touching pair of spheres of radius√

3/(4π)/2. (I mean, of course, when you consider the plots as spherical polar coordinateplots.) For |pz〉, for example, the spheres touch at the z-axis origin and are aligned withthe z-axis: the upper sphere is “positive” and the lower sphere is “negative”: i.e., theradial position from the origin comes out a negative number: one just plots the magnitude.HINTS: It is sufficient to do the proof for |pz〉, since the others are the same mutatis

mutandis. A diagram would help.

b) Interpret the physical significance of the polar plots of the cubical harmonics.

c) Show that |px〉 and |py〉 are eigenstates of L2z, but not Lz. What other angular momentum

operators are they eigenstates of? HINT: Recall

Lz =h−i

∂

∂φ.

d) Now we come to the question yours truly wanted to ask before chronic digression set in.Write sp-bond coupling parameters in terms of fiducial values in units eV-A2: e.g.,

C

d2

A,

where C is a numerical constant (i.e., an actual value) in eV-A2 and dA is mean nuclei

separation in Angstroms. Then evaluate the constants for dA = 3. HINT: Recall

h−2

m= 7.62 eV-A2 ,

where m is the electron mass.

021 qfull 00300 2 3 0 easy math: Li2 with spinExtra keywords: Reference Ha-72

2. Let us consider the single-particle bonding and antibonding states and their energies for Li2.We assume that the single-particle Hamiltonian of the Li2 molecule is

H =−h−2

2m∇2 + V (~r ) =

−h−2

2m∇2 + V (~r − ~r1) + V (~r − ~r2) .

where ~r is measured from the midpoint between the nuclei, ~r1 is the position of nucleus 1,and ~r2 the position of nucleus 2. We are assuming the nuclei are at fixed positions which is the

134 Chapt. 21 Molecules

crudest Born-Oppenheimer approximation. The observed equilibrium separation of the nuclei isd = 2.67 A: we take this to be the fixed separation. We will make the LCAO (linear combinationof atomic orbitals) approximation.

a) Use the linear variational method to calculate the the bonding and antibonding states andtheir energies. The basis states are the two atomic orbital 2s states of the atoms: call them|1〉 and |2〉. We assume |1〉 and |2〉 are knowns: they have the same energy εs = −5.34 eV.Assume the basis states are orthogonal: a poor approximation actually, but this problem isintended to be heuristic. Make a reasonable approximation to evaluate the diagonal matrixelements. For off-diagonal or coupling matrix elements use

Vssσ = −π2

8

h−2

md2= −9.40 eV-A

d2

A

.

this is one of Harrison’s “universal” sp-bond couplings (Ha-95). Which state is the bondingstate and which the antibonding state and why are they so called?

b) What is the component of orbital angular momentum of single-particle states about theinter-nuclear axis? How do you know this? What symbol represents this orbital angularmomentum component value for molecules? HINT: Recall that the z-component orbitalangular momentum operator is

Lz =h−i

∂

∂φ,

where φ is the azimuthal angle about the z-axis (MEL-23).

c) Now construct six plausible symmetrized two-particle states including spinor from ourbonding and antibonding position states: a ground state and the five excited states. Soeveryone is on the same wavelength let

α =

(

10

)

and β =

(

01

)

.

What are the approximate energies of these states? Can we construct any more statesfrom the bonding and antibonding states? We, of course, are assuming that there is nospin operator in the Hamiltonian. NOTE: These states may not be very realistic: this isjust an exercise.

021 qfull 03000 2 5 0 moderate thinking: molecular relative coordinatesExtra keywords: A misconcieved problem.

3. One usually wishes to separate the center of mass and relative parts of the nuclei part of amolecular wave function. For two nuclei, the situation is a two-body problem and can be treatedlike hydrogenic systems (Da-334). For the general multiple nuclei case, a different approach isneeded that treats all nuclei on the same footing. Let ~ri be nucleus i’s position relative toan external inertial frame. Let ~r′i be nucleus i’s position relative to ~R the molecular center ofmass. We make the approximation that the electrons can be neglected in evaluating the centerof mass. NOTE: I was fooled into thinking there was neat way of doing this. The cross termdoesn’t vanish. But one probably has to construct independent coordinates in a special way foreach kind of molecule. But maybe something is salvageable so I’ll leave this around for now.

a) Express ~R and ~r′i in terms of the positions ~ri.

b) Now express the operators ∂/∂xi and ∂2/∂x2i in terms of operators ∂/∂x′i and ∂/∂X.

For simplicity we only consider the x-components of the various coordinates. The y- andz-components are handled similarly.

Chapt. 21 Molecules 135

c) Now show that

H =∑

i

p2i

2Mi+ V (ri) =

p2cm

2Mtot+∑

i

p′2i2M ′

i

+ V (r′i) ,

where H is the Hamiltonian of the nuclei in the effective potential V (ri) (the curlybrackets mean “set of”) with no external potential present, p′i are the relative coordinatemomentum operators, and

M ′i =

Mi

[1 − (Mi/Mtot)]2 .

Chapt. 22 Solids


022 qmult 00100 1 1 3 easy memory: simplest quantum mechanical solid model1. The simplest quantum mechanical solid model is arguably:

a) the hydrogen atom. b) the helium atom. c) the free electron gas model.d) the infinite periodic potential model. e) the finite periodic potential model.

022 qmult 00110 1 1 1 easy memory: infinite square boundary conditions2. For the free electron gas model of a solid, one common simple choice of boundary conditions is:

a) infinite square well boundary conditions. b) finite square well boundary conditions.c) Gaussian well boundary conditions. d) hydrogen atom boundary conditions.e) helium atom boundary conditions.

022 qmult 00130 1 4 4 easy deducto-memory: periodic boundary conditionsExtra keywords: mathematical physics

3. “Let’s play Jeopardy! For $100, the answer is: These quantum mechanical boundary conditionsfor solids, also known a Born-von-Karman boundary conditions, are not realistic in most cases.They are realistic in some cases. For example, for the dimension of a solid that forms a closedloop: e.g., a solid that has donut shape can be have an angular coordinate that must periodic bysymmetry over the range [0, 360]. But whether realistic or not, it can be shown—but no oneever says where—that they lead to the same average behavior as realistic boundary conditionsfor macroscopically large solid samples.

Why are these boundary conditions used at all? Well for one thing they are an ideal kindof boundary conditions that are completely independent of what the surface behavior of solidis. Thus, they are neutral case. For another thing they are easy to use in developments in aparticular when dealing with periodic potentials in a solid.”

What are , Alex?

a) infinite square well boundary conditions b) aperiodic boundary conditionsc) Rabi-Schwinger-Baym-Sutherland boundary conditionsd) periodic boundary conditions e) relaxed boundary conditions

022 qmult 00500 1 4 5 easy deducto-memory: Bloch’s theoremExtra keywords: mathematical physics

4. “Let’s play Jeopardy! For $100, the answer is: It is a theorem in quantum mechanics that appliesto systems with periodic potentials. In one dimension, say one has the periodic potential

V (x) = V (x+ a)

where a is the period distance The theorem then says that the wave function must satisfy

ψ(x+ a) = eKaψ(x) or ψ(x) = eKaψ(x− a)

orψ(x+ na) = eKnaψ(x) or ψ(x) = eKnaψ(x− na) ,

136

Chapt. 22 Solids 137

where K is wave number quantity and n is any integer. Say that one has ψ(x) for the range[0, a]. Then the ψ(x) for the range [na, (n+ 1)a] can be evaluated from the last formula: e.g.,

ψ(na) = eKnaψ(0) and ψ[n+ 1)a] = eKnaψ(a) .”

What is , Alex?

a) Lorentz’s theorem b) Einstein’s rule c) Schrodinger’s last theoremd) Born’s periodicity law e) Bloch’s theorem

022 qmult 00530 1 1 3 easy memory: Dirac comb5. A periodic potential that consists a periodic array of equally strong Dirac delta function

potential spikes separated by flat potential reigons is called a comb.

a) Bloch b) Compton c) Dirac d) Fermi e) Pauli


022 qfull 00100 2 5 0 moderate thinking: free electron solidExtra keywords: (Ha-324:2.4)

1. Let us consider a free electron gas model of a solid in 1, 2, and 3 dimensions simultaneously.Use periodic boundary conditions (AKA Born-Von-Karman boundary conditions) and assumecubical shape in all three cases: a 1-dimensional cube is a line segment, a 2-dimensional cubeis a square, and a 3-dimensional cube is a cube. Let L be the length of a side of the cube andLℓ be the cube volume where ℓ is the number of dimensions.

a) Solve the time-independent Schrodinger equation for the single-particle stationary statesfor all three cases. Normalize the solutions solutions and give their quantization rules forwavenumbers and energy. It is a given that the multi-particle time-independent Schrodingerequation separates into identical single-particle time-independent Schrodinger equations.But it is NOT a given that the single-particle time-independent Schrodinger equationseparates into 3 identical 1-dimensional time-independent Schrodinger equations.

b) What is the volume Vk in k-space of cubes that are centered on the allowed spatial statewave vectors and that are contiguous with each other: i.e., what is the k-space volume perstate? What is the average density of spatial states in k-space ρk?

c) We now make the continuum approximation which is valid for samples that are macroscopicin all available dimensions. This means that we treat the average density of spatial statesin k-space as if it were an uniform density. Find the expression for the differential numberof states dN in a spherical shell in k-space. The shell radius is k and its thickness is dk.Include the spin degeneracy by a factor g which equals 2 for spin 1/2 electrons. But leaveg unevaluated. By leaving g unevaluated, one can track how the spin degeneracy affectsdN and expressions derived from dN .

d) The Pauli exclusion principle for fermions requires that each single particle spatial-spinstate have only one fermion at most. This statement must be qualified. What it reallymeans is that the product wave function of single particle states can have each distinctsingle-particle state included once only. If there are M fermions, the overall symmetrizedwave function containsM ! versions of the product wave function with the individual particlecoordinate labels in all possible permuations. But we don’t have to worry about productwave functions or symmetrized wave functions explicitly in the free electron gas model ofsolids. We simply make use of the Pauli exclusion principle to say that the single-particlestates can only be used once in calculating results or to put this in common, but somewhatmisleading, jargon only one electron can occupy at single particle state at most.

138 Chapt. 22 Solids

Now in the ground state of (which is the absolute zero temperature state) in our freeelectron gas model, the electrons occupy the lowest energy single-particle states consistentwith the Pauli exclusion principle. This means in k-space, the electrons occupy a sphereof radius kF where F where stands for Fermi: since F stands for a name, not a variableit ought to be in Roman, not Italic, font, but convention seems to dictate Italic font (e.g.,Gr-221, CT-1435). The radius kF is called the Fermi wavenumber.

Using the results of the part (c) answer solve for kF for an electron density (in spacespace) of ρe = M/Lℓ.

e) Now solve for the Fermi energy EF for the 3 cases.

f) What is ρE : i.e., the density of states per unit volume per unit energy in the continuum ofstates approximation. Write a general formula that is valid for all three dimension cases.HINT: One requires the same number of states between any corresponding limits: i.e.,

dN = ρk shell dk = ρE dE ,

where ρk shell is the density of spatial-spin states in the differential k space shell dk. Thegeneral expression for ρk shell must have turned up in the part (c) answer without so labelingit.

g) Now solve for the total energy Etotal of the ground state for M electrons, Eave the averageenergy for the M electrons, and Eave/EF . Don’t bother to expand EF using the expressionsfrom the part (e) answer. The formulae for this answer are long enough as it is.

022 qfull 00110 1 3 0 easy math: free electron gas formulae and fiducial formulae2. For a free electron gas (in 3 dimensions) at abolute zero temperature, the Fermi energy is given

by

EF =h−2

2m

[(

2

g

)

(3π2)ρe

]2/3

,

where h− is Planck constant divided by 2π, m is the electron mass, g = 2 is the spin 1/2 particledegeneracy, and

ρe =M

V

is the free electron density with M being the number of electrons and V being the samplevolume. The average energy per electron Eave is given by

Eave =3

5EF .

For this question, you will need the following constants,

e = 1.602176487(40)× 10−19 C ,

m = 9.10938215(45)× 10−31 kg ,

mc2 = 510998.910 eV ,

kB = 1.3806504(24)× 10−23 J/K ,

h− = 1.054571628(53)× 10−34 J, s ,

mamu = 1.660538782(83)× 10−27 kg .

a) Free electron density can be expressed in terms of ordinary density (AKA mass density) ρby

ρe =ρ

mamu

∑

i

XiZi

Ai,


where the sum is over all atoms in the sample, Xi is the mass fraction of atom i, Zi isthe number of free electrons per atom for atom i, and Ai is the atomic weight of atom i.Convince yourself that this formula makes sense. Actually the formula can be simplifiedby introducing the mean mass per electron µe defined by

1

µe=∑

i

XiZi

Ai.

Take 1000 kg/m3 as a fiducial value for ρ (which is just like writing density in grams percubic centimeter). Take 50 as a fiducial value for µe: this is like an element in the atomicweight vicinity of iron with one valence electron delocalized. Now write ρe in terms offiducial values: i.e., find the coefficient in the formula expression

ρe = coefficient × ρ

1000 kg/m3

50

µe,

The coefficient is a fiducial electron density.

b) Now find the formula for EF in terms of fiducial values both for joules and electronvolts:i.e., find the coefficient in the formula

EF = coefficient ×[(

2

g

)

ρ

1000 kg/m3

50

µe

]2/3

.

The coefficient is a fiducial Fermi energy. Are solids in ordinary human environmentsrelativistic?

c) Now find the formula for the Fermi temperature TF = EF /kB in terms of fiducial valuesboth for joules and electronvolts. The coefficient is a fiducial Fermi temperature. Are solidsare ordinary human environments hot or cold in comparison to the Fermi temperature?

d) Now find the formula for the Fermi velocity vF =√

2EF /m (which is the non-relativisticformula) in terms of fiducial values. The coefficient is a fiducial Fermi velocity.

e) For zero temperature,

P = −∂E∂V

where E is the total energy of the sample and V is the sample volume. A free electrongas exhibits a pressure derivable from this classical expression. The pressure is called thedegeneracy pressure and it’s equation of state is quite unlike that for an ideal gas. Derivethe zero-temperature free-electron-gas pressure formula as a function of ρe. Then find thepressure formula in terms of fiducial values.

f) The bulk modulus of a material is a measure of its incompressibility or stiffness. Thedefinition is

B = −V dPdV

= ρdP

dρ

where temperature needs to be specified in general and were we have used

dV = d

(

1

ρ

)

= − 1

ρ2dρ = −V

ρdρ

which can also be writtendV

V= −dρ

ρ.


One can see that bulk modulus is a measure stiffness by rewriting to

dρ

ρ=dP

B.

The bulk modulus is the coefficient needed to convert a differential change in pressure in thecorresponding RELATIVE differential change in density. The bigger the bulk modulus,the stiffer the substance.

Find the bulk modulus formula for the zero-tempature free electron gas. Then findthe bulk modulus formula in terms of fiducial values.

022 qfull 00200 2 5 0 moderate thinking: computing Fermi energiesExtra keywords: (Ha-324:2.3) free electron metals

3. The metals Na, Mg, and Al have, respectively 1, 2, and 3 free electrons per atom and volumesper atom 39.3 A3, 23.0 A3, and 16.6 A3. At zero temperature what is the Fermi energy to whichthe free electron states are filled?

022 qfull 00500 1 3 0 easy math: energy band structure functionExtra keywords: Bloch states

4. The energy band structure of the 1-dimensional Dirac comb solid is determined by the equation

cos(Ka) = cos(z) +β

zsin(z) ,

where a is the cell size, K is the Bloch wavenumber, z = ka ≥ 0 (where k is the wavenumberfor a single cell), and

β =mαa

h−2

(where m is electron mass and α is the strength of the Dirac delta function potentials that makeup the Dirac comb). The Bloch wavenumbers are quantized according to the rule

Ka =2π

Nm ,

where m = 0, 1, 2, 3, . . . , N−1 (where N is the number of cells). Actually, Ka is not determinedto within an additive constant of 2π. Therefore one could also specify the range for m as

m = −N2,−N

2+ 1, . . . ,

N

2− 1,

N

2

since for N large one can with negligible error approximateN as even no matter what it actuallyis and add on an extra unphysical state. This second specification for m is more symmetricaland, as it turns out, makes k vary monotonically with K within a band. For definiteness in thisproblem, we assume β ≥ 0.

The only possible solution z values are those that make confine

f(z) = cos(z) +β

zsin(z)

to the range [−1, 1]. For a Ka value, one solves

cos(Ka) = f(z)

for ka and then k. Thus, the k values have a non-trivial quantization rule. The energy of thek state is then determined by

E =h−2k2

2m.


The energies have a non-trivial quantization rule too.

A key point is the function f(z) can oscillate above 1 and below −1 and in those regionsthere are no solutions for k, and therefore no allowed states and no solutions for E. Thoseenergy regions are the famous energy gaps. The allowed regions are the energy bands. Since Nis an enormous number for macroscopic samples, the energy bands are very dense in number ofstates and one gets states virtually to the z points where |f(z)| = 1. The z values that yield|f(z)| = 1 are the band limits. Each band has a start and end limit. There are correspondingenergy start and end limits.

a) Determine the formula end limit of a general ℓth energy band. Note ℓ = 1, 2, 3, . . . Showthat the formula is the end limit formula and never gives start limits. HINT: Finding theformula is easy. Showing that it is the end limit requires taking the derivative of f(z).

b) A key point to proof is that there is only one limit point between two consecutive end limitpoints given by the formula in the part (a). This single limit point is a start limit point.

In a test situation just assume there is a single start limit point between the end limitpoints and go onto the rest of the problem.

In a non-test situation, do the proof. HINT: Prove there is a single stationary pointbetween the consecutive end limit points. The function f(z) value at the stationary pointis must be outside of the range [−1, 1] by the part (a) answer. Also from the part (a)answer, the function at the stationary point must be greater than 1 if the next end limitpoint f(z) = −1 and the function at the stationary point must be less than 1 if the nextend limit point is f(z) = 1 Thus from the stationary point to the next end limit pointthe function is is monotonic and crosses into the allowed range once only. That is crossingis the start limit point itself. The stationary point itself can’t be found analytically ingeneral, but its uniqueness can be proven by arranging the equation for stationary pointinto a form where tan(z)/z equals something.

c) Find an approximate formula for the start limit for the first band in the limit that β isvery large. Show that this formula also gives the start limit for β = 0, and thus constitutesan interpolation formula: a formula that gives correct limiting behavior and interpolatessmoothly in between those limits. HINT: Set the function f(z) equal to 1 and expand thetrigonometric functions in f(z) about π to 1st order in small z−π. Why are the expansionsthe good thing to do and why is it good not to expandi the 1/z factor to 1st order in smallz − π even though it would be consistent with the other expansions?

d) Find an approximate formula for the start limit ∆z = z−zℓ−1 = z−(ℓ−1)π for the generalℓth band, except that ℓ ≥ 1. HINT: Set f(z) = (−1)ℓ−1 and expand cos(z) to second orderin small ∆z about zℓ−1. and sin(z) to 3rd order in small ∆z about zℓ−1 = (ℓ−1)π Do NOTapproximate 1/(zℓ−1 + ∆z. Write β = yzℓ−1/2 as a simplification. What approximate sizelimit must be put on ∆z for our approximations to be valid? What is ∆z in the limit thatzℓ−1 becomes very large.

e) Take the interpolation formula from the part (c) answer the ∆z formula for very large zℓ−1

from the part (d) answer and invent an interpolation formula for ∆z valid for all ℓ. Wemean it is valid for ℓ in that it gives the right limiting behavior for β for the ℓ = 1 and theright limiting behaviro for ℓ becoming very large. Since it has those right limiting behaviors,it may well interpolate to some accuracy everywhere. HINT: There is no absolutely rightanswer, but a fairly obvious interpolation formula leaped to yours truly’s eye.

022 qfull 00510 1 3 0 easy math: energy band structure function 2

5. This a super-easy problem if you can understand the lengthy setup.

For the 1-dimensional Dirac comb potential, the band structure is determined by theequation

cos(Z) = f(z) ,


where

f(z) = cos(z) +β

zsin(z) .

The Z is determined by the Bloch wavenumber quantization rule

Z = Ka =2π

Nm where m = 0, 1, 2, . . . , (N − 1)

where N is the number of number of cells and a is the cell length. For a macroscopic systemN is huge, and so Z can be close to any real number in the range [0, 2π]. The z = ka, where kis the ordinary wavenumber from which the energy of a single-particle state can be determinedfrom

E =h−2k2

2m.

If we knew a z value, determining the corresponding Z value would be a cinch. But weknow the Z values and need to determine the z values. There is no analytic solution for generalz.

We do know that the only solutions are in bands set by the fact that cos(Z) can only bein the range [−1, 1]. From earlier work, we know that

zℓ,start = (ℓ− 1)π + ∆zℓ = (ℓ− 1)π +2βπ

2β + 4 + π2(ℓ− 1)

is a good approximation for the start limit of band ℓ and

zℓ,end = ℓπ

is the exact result for the end limit of a band ℓ. The band numbers run form ℓ = 1 to ℓ = ∞.The cos(Z) value sweeps from 1 to −1 as z increases for odd bands and from −1 to 1 for evenbands.

Let’s define a parameter g that runs from 0 to 1 and parameterize Z in terms of g thusly

Z =

gπ for ℓ odd;(g + 1)π for ℓ even.

Using g (which we can easily find for any Z value and any ℓ) and the formulae for zℓ,start

and zℓ,end devise an approximate z formula that is LINEAR in g. HINT: This is super-easy,but you should test that g = 0 and g = 1 give the correct band limits: z(ℓ, g = 0) = zℓ,start andz(ℓ, g = 1) = zℓ,end.

Chapt. 23 Interaction of Radiation and Matter


023 qmult 00100 1 1 4 easy memory: Maxwell’s equations1. Classical electrodynamics is summarized in:

a) Stirling’s formula.b) Heitler’s five equations.c) Montefeltro’s three laws.d) Maxwell’s four equations.e) Euclid’s axioms.

023 qmult 00200 1 4 5 easy deducto-memory: EM potentials2. “Let’s play Jeopardy! For $100, the answer is:

~E = −∇Φ − 1

c

∂ ~A

∂tand ~B = ∇× ~A .”

a) What are Maxwell’s equations, Alex?b) What are two arbitrary equations, Alex?c) What are the gauge transformation equations, Alex?d) What are the Noman equations, Alex?e) What are the equations for deriving the electromagnetic fields from the electromagnetic

potentials, Alex?

023 qmult 00300 1 1 3 easy memory: transverse gauge3. The condition imposed for the transverse gauge (or radiation or Coulomb gauge) is:

a) ∇ · ~B = 0.

b) ∇ · ~A+ (1/c)∂Φ/∂t = 0.

c) ∇ · ~A = 0.

d) ~A′ = ~A+ ∇Λ.e) Φ′ = Φ − (1/c)∂Λ/∂t.

023 qmult 00400 1 4 2 easy deducto-memory: EM energy densityExtra keywords: Reference Ja-236

4. “Let’s play Jeopardy! For $100, the answer is:

E =E2 +B2

8π.”

a) What Laplace’s equation, Alex?b) What the energy density of the electromagnetic field in SI units, Alex?c) What the energy density of the electromagnetic field in Gaussian units, Alex?d) What Poynting’s vector, Alex?

143

144 Chapt. 23 Interaction of Radiation and Matter

e) What is an example of using “E” for two quantities in one equation, Alex?

023 qmult 00500 1 1 2 easy memory: wave equation solution5. The simple wave equation

∇2~g =1

c

∂2~g

∂t2

has a very general solution of the form:

a) ~g = 0.

b) ~g(~k · ~r − ωt), where ~g is any vector function and ω = |~k|c.c) ~g(~k × ~r − ωt), where ~g is any vector function and ω = |~k|c.d) ~g(ωt).

e) ~g(~k · ~r).

023 qmult 00600 1 1 2 easy memory: EM perpendicular vectors6. For self-propagating electromagnetic radiation the electric field, magnetic field, and propagation

directions are:

a) collinear.b) mutually perpendicular.c) self-perpendicular.d) arbitrary.e) random.

023 qmult 00700 1 4 4 easy deducto-memory: box quantization7. “Let’s play Jeopardy! For $100, the answer is: A standard trick in quantum mechanics for

replacing basically isotropic, homogeneous systems with complex boundary conditions by asimpler, tractable, periodic-boundary-condition system which—according to the faith—musthave the same bulk properties.”

a) What is nothing I’ve ever heard of, Alex?b) What is the tethered function method, Alex?c) What is black magic, Alex?d) What is box quantization, Alex?e) What is the way to get a completely wrong answer, Alex?

023 qmult 00800 1 4 3 easy deducto-memory: interaction Hamiltonian8. “Let’s play Jeopardy! For $100, the answer is:

H =1

2m

(

~p− e

c~A)2

+ eΦ + V .”

a) What is Schrodinger’s equation, Alex?b) What is the Hamiltonian for the simple harmonic oscillator, Alex?c) What is the particle Hamiltonian including the interaction with the electromagnetic field,

Alex?d) What is the interaction Hamiltonian of the electromagnetic field, Alex?e) What is any old Hamiltonian, Alex?

023 qmult 01200 1 1 4 easy memory: cross section9. What one often wants from a radiation interaction calculation is a coefficient or the like that

can be employed in macroscopic radiative transfer: e.g.,

a) a postulate.

Chapt. 23 Interaction of Radiation and Matter 145

b) a postulant.c) a golden section.d) a cross section.e) a crossbow.

023 qmult 01300 1 1 1 easy memory: quantizing radiation10. To quantize the electromagnetic radiation field we assert that the energy of radiation mode

cannot take on a continuum of values, but must come in:

a) integral amounts of a basic unit called a quantum of electromagnetic radiation or a photon.b) amounts related by the golden section.c) photoids.d) integral numbers of ergs.e) integral numbers of eVs.

023 qmult 01400 1 1 3 easy memory: creation and annihilation operators11. The electromagnetic field operator is defined in terms of:

a) Bell and Sprint operators.b) left and right operators.c) creation and annihilation operators.d) procreation and inhibition operators.e) genesis and revelations operators.

023 qmult 01500 1 1 2 easy memory: zero point energy12. In the formalism of quantized electromagnetic radiation, the fact that the creation and

annihilation operators don’t commute leads to:

a) nothing in particular.b) the existence of zero-point energy.c) energy less than the zero-point energy.d) living close to the office.e) living in the office.

023 qmult 01600 1 4 4 easy deducto-memory: Einstein stimulated em.13. “Let’s play Jeopardy! For $100, the answer is: An effect discovered by Einstein by means of a

thermodynamic detailed balance argument.”

a) What is spontaneous emission, Alex?b) What is special relativity, Alex?c) What is the photoelectric effect, Alex?d) What is stimulated emission, Alex?e) What is spontaneous omission, Alex?

023 qmult 01700 1 4 5 easy deducto-memory: free particles don’t radiate14. “Let’s play Jeopardy! For $100, the answer is: A particle that cannot radiate by a Fermi golden

rule process in the non-relativistic limit and perhaps not at all—but one never knows whatspecial exotic cases may be said to allow it to radiate.”

a) What is a classical particle, Alex?b) What is a neutral pion, Alex?c) What is a quark, Alex?d) What is an unbound particle subject to potentials, Alex?e) What is a free particle, Alex?


023 qmult 01800 1 1 1 easy memory: conservation of momentum15. Since it seems that no quantum mechanical matter particle can actually be in a pure momentum

eigenstate, conservation of momentum in radiative emission or absorption in the instructor’sview

a) cannot be strict in a classical sense.b) can be strict in a classical sense.c) both can and cannot be strict in a classical sense.d) can neither be nor not be strict in a classical sense.e) is moot and categorically valid in a classical sense.

023 qmult 01900 1 1 2 easy memory: electric dipole transitions16. Typically, strong atomic and molecular transitions are

a) electric quadrupole transitions.b) electric dipole transitions.c) magnetic dipole transitions.d) electric monopole transitions.e) magnetic metropole transitions.

023 qmult 02000 1 4 4 easy deducto-memory: selection rules17. “Let’s play Jeopardy! For $100, the answer is: The selection rules for electric dipole

transitions (i.e., allowed transitions) between energy eigenstates whose angular parts arespherical harmonics.”

a) What are Hund’s rules, Alex?b) What are ∆m = ±2 and ∆ℓ = 0, Alex?c) What are ∆m = ±1 and ∆ℓ = 0 or ±1, Alex?d) What are ∆m = 0 or ±1 and ∆ℓ = ±1, Alex?e) What are taking the fresher and firmer ones, Alex?


023 qfull 00300 2 5 0 moderate thinking: classical EM scatteringExtra keywords: reference Mi-83

1. Say we had a classical simple harmonic oscillator (SHO) consisting of a particle with mass mand charge e and a restoring force mω2

0 where ω0 is the simple harmonic oscillator frequency.This SHO is subject to driving force caused by traveling electromagnetic field (i.e., light):

~Fdrive = e ~E0eiωt ,

where ~E0 is the amplitude, ω is the driving frequency, and we have used the complex exponentialform for mathematical convenience: the real part of this force is the real force. The magneticforce can be neglected for non-relativistic velocities. The Lorentz force is

~F = e

(

~E +~v

c× ~B

)

(Ja-238) and ~E and ~B are comparable in size for electromagnetic radiation, and so the magneticforce is of order v/c smaller than the electric force. (See also MEL-130.) An oscillating charge isan accelerating charge and will radiate electromagnetic radiation. The power radiated classicallyis

P =2e2a2

3c3,


where ~a is the charge acceleration. This radiation causes an effective damping force givenapproximately by

~Fdamp = −mγ~v ,where

γ =2e2ω2

0

3mc3.

The full classical equation of motion of the particle is

m~a = −mω20~r + e ~E0e

iωt −mγ~v .

a) Solve the equation of motion for ~r and ~a. HINTS: The old trial solution approach works.Don’t forget to take the real parts although no need to work out the real part explicitly:i.e., Re[solution] is good enough for the moment.

b) Now solve for the time average of the power radiated by the particle. HINT: You willneed the explicit real acceleration now.

c) The average power radiated must equal the average power absorbed. Let’s say that theparticle is in radiation flux from a single direction with specific intensity

I0 =cE2

0

8π

(Mi-9), where time averaging is assumed como usual. The power absorbed from this fluxis σ(ω)I0, where σ(ω) is the cross section for energy removed. Solve for σ(ω) and thenfind show that it can be approximated by a Lorentzian function of ω with a coefficientπe2/(mc). HINT: It is convenient to absorb some of the annoying constants into anotherfactor of γ.

d) Now rewrite the cross section as a function of ν = ω/(2π) (i.e., the ordinary frequency)and then integrate over ν to get the frequency integrated cross section σν int of the system.What is the remarkable thing about σν int? Think about how it relates to the systemfrom which we derived it. Evaluate this frequency integrated cross section for an electron.HINT: The following constants might be useful

α =e2

h−c=

1

137.036, h− = 1.05457× 10−27 erg s , and me = 9.10939× 10−28 g .

023 qfull 00500 2 5 0 moderate thinking: gauge invarianceExtra keywords: (Ba-299:1)

2. The gauge transformations of the electromagnetic potentials are:

~A(~r, t)′ = ~A(~r, t) + ∇χ(~r, t)

and

φ(~r, t)′ = φ(~r, t) − 1

c

∂χ(~r, t)

∂t

(Ja-220), where ~A(~r, t) is the vector potential, φ(~r, t) is the scalar potential, the primedquantities are the transformed versions, and χ(~r, t) is the gauge. Show that the solution tothe time-dependent Schrodinger equation for n particles of charge e and mass m transforms so

Ψ(~r1, . . . , ~rn, t)′ = exp

[

ie

h−c

n∑

i=1

χ(~ri, t)

]

Ψ(~r1, . . . , rn, t)


under a gauge transformation of the potentials treated as classical fields: i.e., if Ψ′ satisfies aprimed version of the Schrodinger equation, then Ψ satisfies an unprimed version. HINT: Itsuffices to consider one particle in one dimension. I know of no simplifying tricks. You justhave to grind out the 15 odd terms very carefully.

023 qfull 00700 2 5 0 moderate thinking: QM CA operatorsExtra keywords: (Ba-299:3)

3. The photon creation and destruction operators of the electromagnetic field photons (the CAoperators for short) are defined by, respectively

A†~k~λ

| . . . , N~k~λ, . . .〉 =

√

2πh−c2ω

√

N~k~λ + 1| . . . , N~k~λ + 1, . . .〉

and

A~k~λ| . . . , N~k~λ, . . .〉 =

√

2πh−c2ω

√

N~k~λ| . . . , N~k~λ − 1, . . .〉 ,

where N~k~λ is the number of photons in the box quantization mode specified by wavenumber~k and polarization vector ~λ. All the physically unique states created by these operators areassumed to be orthogonal. From this assumption it follows that creation and annihilationoperators are Hermitian conjugates as their labeling indicates: we’ll leave that proof sine die.The annihilation operator acting on the empty mode (i.e., one with N~k~λ = 0) gives a zero ornull state (or vector). Any operator acting on a zero state gives a zero state, of course. Thus,non-zero commutator identities can’t be proven by acting on zero states and it is understoodthat zero states are never thought of in proving commutator identities.

Dimensionless forms of these operators are, respectively,

a†i | . . . , Ni, . . .〉 =√

Ni + 1| . . . , Ni + 1, . . .〉and

ai| . . . , Ni, . . .〉 =√

Ni| . . . , Ni − 1, . . .〉 ,

where index i subsumes both the ~k and ~λ indices.

a) Prove the commutator identities

[ai, aj ] = 0 , [a†i , a†j] = 0 and [ai, a

†j] = δij .

Remember the case of i = j for the first two identities. What are the dimensioned formsof these commutator identities?

b) Recall that classically the total energy in the radiation field is

E =

∫ (

E2 +B2

8π

)

d~r ,

where calligraphic E is total energy, E is the electric field, B is the magnetic field (ormagnetic induction if you prefer), and the integration is implicitly over all the volume ofthe system. The quantum mechanical, Heisenberg representation field operator for theradiation field is

~Aop =∑

~k~λ

[

A~k~λ~λei(~k·~r−ωt)

√V

+ H.C.

]

,

where we have assumed box quantization with periodic boundary conditions, V is thevolume of the box, and “H.C.” stands for Hermitian conjugate. What is the time-averaged


Heisenberg representation of the Hamiltonian of the electromagnetic field Hem in terms ofthe operator A†

~k~λA~k~λ?

c) What is Hem in terms of dimensionless EM operators? What is the zero point energy of aradiation field: i.e., the energy expectation value for the state |0, 0, 0, . . . , 0〉?

d) Convert the summation for the zero point energy into an integral using the continuumapproximation for the box quantization. What is the energy density E0/V of the vacuumup to the mode with photon energy ǫ? (Note that ǫ is not the energy in the mode, it isthe energy of photons in the mode: each photon has ǫ = h−ω.) What unfortunate thinghappens if you set ǫ = ∞?

023 qfull 00800 2 5 0 moderate thinking: commutation field operatorsExtra keywords: (Ba-299:4)

4. The quantum mechanical, Heisenberg representation field operator for the radiation field is

~Aop(~r, t) =∑

~k~λ

[

A~k~λ~λei(~k·~r−ωt)

√V

+ H.C.

]

,

where we have assumed box quantization with periodic boundary conditions, V is the volumeof the box, and “H.C.” stands for Hermitian conjugate.

a) Determine expressions for ~Eop(~r, t) and ~Bop(~r, t).

b) Determine [ ~Eop(~r, t), ~Bop(~r, t)]. What can you say about the simultaneous knowledge that

one can have about ~E and ~B? HINTS: The following commutator relations will helpsimplify:

[A~k~λ, A~k′~λ′] = 0 , [A†

~k~λ, A†

~k′~λ′] = 0 , and [A~k~λ, A

†~k′~λ′

] = δ~k~k′δ~λ~λ′

2πh−c2ω

,

where ω = kc, of course. You will also have to deal with the outer product of two vectorswhich is a tensor: e.g., ~a⊗~b. The outer product is commutative: e.g., ~a⊗~b = ~b ⊗ ~a (e.g.,ABS-25).

023 qfull 01500 2 5 0 moderate thinking: free particles non-radiateExtra keywords: Reference Ba-280

5. Say you have a free particle (i.e., a free matter particle) in a momentum eigenstate with eigenmomentum h−~qn. Remember a free particle in the conventional sense of quantum mechanicsmeans one not affected by any potentials at all. Using Fermi’s golden rule you can try tocalculate the particle’s spontaneous emission when making a transition to another momentumeigen state with eigen momentum ~q0 (Ba-279). You find a momentum conservation rule isimposed:

h−~k = h−~qn − h−~q0 ,where h−~k is the momentum of the emitted photon. But since you used the golden rule you alsoimposed energy conservation:

h−ck =h−2q2n

2m− h−2

q202m

,

where m is the particle’s mass. The conclusion that is implied by these two relations is that freeparticles can’t radiate at least not through the golden rule process used in the calculations. Showhow the conclusion is reached. HINT: The two relations actually can be mutually satisfied ina mathematical sense: think about the assumptions implicit in them.


023 qfull 02000 3 5 0 moderate thinking: hydrogen spontaneous emissionExtra keywords: (Ba-300:7)

6. Consider a hydrogen atom with the nucleus fixed in space.

a) What are all the states of the two lowest principal quantum numbers in spectroscopicnotation: e.g., 2s(m = 0) for n=2, ℓ=0 (which is what the s symbol means), and m = 0.Don’t distinguish states by spin state.

b) Between which states are there (electric dipole) allowed transitions transitions.

c) The spontaneous emission power per solid angle in polarization ~λ in the electric dipoleapproximation is

dPλ

dΩ=ω4e2

2πc3|~d0n · ~λ∗|2 ,

where~d0n = 〈0|~R|n〉

is the dipole moment matrix element, |n〉 and |0〉 are, respectively, the initial and finalstates, and

~R =∑

i

ri

is the position operator for all the particles in the system (Ba-282). Find the expression

for ~d0n between the 1s(m = 0) and 2p(m = 0) levels of hydrogen.

023 qfull 02500 2 5 0 moderate thinking: Einstein A coefficientExtra keywords: See Ba-282, Gr-311–313

7. In the electric dipole approximation, the power in polarization λ per unit solid angle per particle(or system of particles) of a spontaneous transition process is

dPλ

dΩ=ω4e2

2πc3|~d0n · ~λ∗|2 ,

where e is the charge on the particle (or particles),

~d0n = 〈0|~R|n〉

is the off-diagonal element of the dipole moment operator, and n and 0 label the initial andfinal states, respectively (Ba-282). In general,

~R =∑

i

~ri .

where the sum is over the position operators of the particles in the system.

a) Consider an ordinary 3-d Cartesian set of axes and radiation emitted along the z-axis.Consider the polarization vectors to be unit vectors in the x and y directions. The dipolematrix element points in an arbitrary direction (θ, φ) in spherical polar coordinates. Whatis the expression for power SUMMED over the two polarization directions in terms of theangle coordinates of the dipole matrix element? HINTS: A diagram probably helps andknowing how to express ~d0n along Cartesian coordinates, but in spherical polar coordinates.

b) Using the expression from the part (a) answer, find the total power radiated by integratingover all solid angle. HINT: Nothing forbids you from mentally transposing the z-axis fromthe general direction to a convenient direction.


c) What is the total photon number rate emitted from the transition (i.e., the Einstein Acoefficient at least as Gr-312 defines it)?

d) If at time zero you have set of N0 atoms in an excited state with only one downwardtransition available with Einstein A coefficient A and spontaneous emission as the onlyprocess, what is the population N at any time t > 0?

Chapt. 24 Second Quantization


024 qmult 00100 1 1 2 easy memory: 2nd quantization1. The formalism for quantizing fields is called:

a) first quantization.b) second quantization.c) third quantization.d) fourth quantization.e) many quantization.

024 qmult 00200 1 4 3 easy deducto-memory: CA anti/commutation2. “Let’s play Jeopardy! For $100, the answer is: These operators for a single mode have

commutation relation [a, a†] = 1 for bosons and anticommutation relation a, a† = 1 forfermions.”

a) What are Hermitian operators, Alex?b) What are anti-Hermitian operators, Alex?c) What are creation and annihilation operators, Alex?d) What are penultimate and antepenultimate operators, Alex?e) What are genesis and revelations operators, Alex?

024 qmult 00300 1 1 1 easy memory: boson CA operator effects3. For bosons the operators a and a† acting on a state |n〉 give, respectively:

a)√n|n− 1〉 (or the null vector if n = 0) and

√n+ 1|n〉.

b)√n− 1|n− 1〉 (or the null vector if n = 0) and

√n|n〉.

c)√n| − n+ 1〉 (or the null vector if n = 0) and

√−n| − n+ 1〉.

d) the null vector and the infinite vector.e) a non-vector and a non-non-vector.

024 qmult 00400 1 1 4 easy memory: fermion number operator4. What are the eigenvalues of the fermion number operator obtained from the matrix

representation this operator: i.e., from

N =

(

0 00 1

)

?

a) ±1b) Both are zero.c) 1 and 2.d) 0 and 1.e) ±2.

152

Chapt. 24 Second Quantization 153

024 qmult 00500 1 1 1 easy memory: number operators commute5. The number operators for different modes for the boson and fermion cases:

a) commute.b) anticommutec) commute and anticommute, respectively.d) anticommute and commute, respectively.e) promote and demote, respectively.

024 qmult 00600 1 1 2 easy memory: CA operators for symmetrization6. Symmetrized states in second quantization formalism are easily constructed using

a) annihilation operators.b) creation operators.c) more annihilation operators than creation operators.d) steady-state operators.e) degenerate operators.

024 qmult 00700 1 4 5 easy deducto-memory: field operators7. “Let’s play Jeopardy! For $100, the answer is: These operators are constructed from the creation

and annihilation operators in second quantization formalism.”

a) What are the copse operators, Alex?b) What are the glade operators, Alex?c) What are the meadow operators, Alex?d) What are the field marshal operators, Alex?e) What are the field operators, Alex?

024 qmult 01000 1 4 4 easy deducto-memory: 2nd QM density operatorExtra keywords: Reference Ba-422

8. “Let’s play Jeopardy! For $100, the answer is:

ρ(~r ) = Ψs(~r )†Ψs(~r ) ,

where Ψs(~r )† and Ψs(~r ) are the field creation and annihilation operators at a point ~r with spincoordinate s.”

a) What is a non-Hermitian operator, Alex?b) What is the density expectation value, Alex?c) What is the first quantization density operaty, Alex?d) What is the second quantization particle density operator, Alex?e) What is it’s Greek to me, Alex?

024 qmult 01010 1 1 1 easy memory: density operator expectation valueExtra keywords: See Ba-422

9. There is a second quantization operator

ρ(~r ) = Ψs(~r)†Ψs(~r ) ,

where Ψs(~r )† and Ψs(~r ) are the field creation and annihilation operators at a point ~r with spincoordinate s. Its expectation value is:

a) the mean density of particles per unit volume.b) the normalization constant of the state |Φ〉 IT is applied to.c) the pair correlation function.

154 Chapt. 24 Second Quantization

d) the pair anti-correlation function.e) the one-particle density matrix.

024 qmult 01100 1 1 3 easy memory: zeros of one-particle density matrixExtra keywords: See Ba-426

10. The one-particle density matrix for non-interacting, spin 1/2 fermions in the ground state of abox quantization system is

Gs(~r − ~r ′) =3n

2

sin(x) − x cos(x)

x3,

where n is the expectation particle density of the ground state and x = pf |~r − ~r ′| (i.e., Fermimomentum times displacement vector). The zeros of this function are approximately given by:

a) x = n, where n = 0, 1, 2, . . .b) x = nπ, where n = 1, 2, 3, . . .c) x = (1/2 + n)π, where n = 1, 2, 3, . . .d) x = (1/2 + n)π, where n = 0, 1, 2, . . .e) x = nπ, where n = 0, 1, 2, . . .

024 qmult 01300 1 1 2 easy memory: pair correlation functionExtra keywords: See Ba-429

11. For a box quantization system of non-interacting, spin 1/2 fermions in the ground state we havethe useful function

gss′(~r − ~r ′) =

1 for s 6= s′;

1 − 9

x6[sin(x) − x cos(x)]2 for s = s′.

where and x = pf |~r − ~r ′| (i.e., Fermi momentum times displacement vector). This function isthe:

a) pair anti-correlation function.b) pair correlation function.c) pair annihilation function.d) pair creation function.e) one-particle density matrix.

024 qmult 01700 1 4 3 easy deducto-memory: exchange effect12. “Let’s play Jeopardy! For $100, the answer is: This effect generally decreases the absolute value

of the potential energy of an interaction between fermions of the same spin coordinate—exceptin the unusual case that the interaction increases with INCREASING distance between thefermions.”

a) What is expunge effect, Alex?b) What is the interchange effect, Alex?c) What is the exchange effect, Alex?d) What is the interstate effect, Alex?e) What is the exchange rate effect, Alex?

024 qmult 02000 1 1 2 easy memory: Feynman diagramsExtra keywords: See Ha-211

13. As a visualization of the terms in an interaction perturbation series in second quantizationformalism, one can use:

a) Feynman landscapes.


b) Feynman diagrams.c) Feynman water colors.d) Feynman lithographs.e) Feynman doodles.


024 qfull 00100 2 5 0 moderate thinking: boson CA proofs1. Let’s do proofs with the boson creation and annihilation (CA) operators a and a† for a single

mode. Recall [a, a†] = 1 is their fundamental commutation relation.

a) The number operator is defined by N = a†a. Show that N is Hermitian. Assume N isan observable (i.e., a Hermitian operator with a complete set of eigenstates for whateverspace we are dealing with) and let its eigen-equation be

N |n〉 = n|n〉 .

We also assume there is no degeneracy. Show that n ≥ 0 and that a|0〉 = 0 (i.e., a|0〉 is thenull vector). HINT: Make use of the rule—which I think is valid—at least lots of sourcesuse it or so I seem to recall—that a legitimate operator acting on a vector always yields avector.

b) Find explicit expressions for [N, a] and [N, a†].

c) Show that a†|n〉 and a|n〉 are eigenstates of N and find explicit expressions for them.

d) Show that n must be an integer.

e) Find a general expression for |n〉 in terms of the vacuum state |0〉. Make sure that |n〉 isproperly normalized.

f) Since the |n〉 are non-degenerate states of the number operator N = a†a (an observable),they are guaranteed to be orthogonal. But for the sake of paranoia vis-a-vis the universe,show explicitly that

〈m|n〉 = δmn ,

making use of the part (e) answer and assuming the vacuum state is properly normalized.HINT: This is easy after you’ve seen the trick.

g) Prove [a, (a†)n] = n(a†)n−1 for n ≥ 1.

024 qfull 00200 2 5 0 moderate thinking: 2 mode fermion CA operatorsExtra keywords: (Ba-439:1)

2. Let us consider the two-mode fermion case.a) Construct explicit 4 × 4 matrix representations of the creation and annihilation (CA)

operators a0, a†0, a1, and a†2. HINT: See Ba-414–415.

b) Construct explicit 4 × 4 matrix representations of the number operators N0, N1, andN = N0 + N1. What are the eigenvalues and eigenvectors the of number operators?Are there any degeneracies?

c) Confirm that the all commutation relations given on Ba-416 hold for the CA operators inthe matrix representation. HINT: I gave up after doing

a0, a†0 = 1 and a0, a

†1 = 0 .


024 qfull 00300 2 5 0 moderate thinking: one-particle density matrixExtra keywords: See Ba-425

3. A quantity that turns out to be useful in studying non-interacting, spin 1/2 fermions in a boxquantization system of volume V is the one-particle density matrix defined by

Gs(~r − ~r ′) = 〈Φ0|Ψs(~r )†Ψs(~r′)|Φ0〉 ,

where |Φ0〉 is the ground state and

Ψs(~r )† =∑

~p

e−i~p·~r

√V

a†~ps and Ψs(~r ) =∑

~p

ei~p·~r

√Va~ps

are the field creation and annihilation operators with h− set to 1 and where s is the z-quantumnumber for the spin state. Recall that the ground state has all the states occupied for |~p | ≤ pf ,where pf is the Fermi momentum. to

a) Find an expression for Gs(0) using the approximation of a continuum of states. What, infact, is this quantity?

b) Show that Gs(~r − ~r ′) is given by

Gs(~r − ~r ′) =3n

2

sin(x) − x cos(x)

x3,

where x ≡ pf |~r−~r ′|, n is the mean density, and the approximation of a continuum of stateshas been used. HINT: The integrand is not isotropic in this case, but one can choose thez-axis for maximum simplicity.

c) Now let us analyze the dimensionless Gs(~r − ~r ′) given by

g(x) =sin(x) − x cos(x)

x3,

where only x greater than is meaningful, of course. First, what is the variation in x usuallyto be thought to be attributed to? Second, what are the small x and large x limiting formsof g(x): give the small x limiting form to 4th order in x. Third, give an approximateexpression for the zeros of g(x). Fourth, sketch g(x).

d) Now find a convergent iteration formula that allows you to solve for the zeros of g(x).Implement this formula in a computer code and compute the first ten zeros to goodaccuracy.

024 qfull 00400 2 5 0 moderate thinking: fermion pair correlationExtra keywords: See Ba-428–429 and Ar-527

4. The pair correlation function for spin 1/2 fermions in the ground state of a box quantizationsystem is

gss′(~r − ~r ′) =

1 , for s 6= s′;

1 − 9

x6[sin(x) − x cos(x)]2 , for s = s′,

where s is a spin coordinate label, x = pf |~r−~r ′|, pf is the Fermi momentum, ~r is the point wherea fermion has be removed, and ~r ′ is the point where one has measured the particle density justafter the removal of the fermion (Ba-428–429). The expectation particle density density for agiven is spin coordinate is (n/2)gss′(~r−~r ′), where n = N/V is the original expectation densityof a spin state: N is the number of particles in the system and V is the system volume.


a) Prove that (n/2)gss′(~r − ~r ′) is properly scaled for s 6= s′: i.e., that the integral of(n/2)gss′(~r − ~r ′) over all volume is the right number of particles.

b) Now prove that (n/2)gss′(~r − ~r ′) is properly scaled for s = s′. HINT: You might haveto look up the properties of the spherical Bessel functions—unless you know those like theback of your hand.

024 qfull 00450 2 5 0 moderate thinking: 2nd quantization potentialExtra keywords: See Ba-434

5. The 2nd quantization operator for a two-body potential V (~r−~r ′) between identical particles is

V2nd =1

2

∑

s′s

∫

d~r ′ d~r V (~r − ~r ′)Ψs(~r )†Ψs′(~r ′)†Ψs′(~r ′)Ψs(~r ) ,

where the 1/2 prevents double counting in the integration, the sum is over all spin states, andthe Ψ†’s and Ψ’s are the field creation and annihilation operators. Prove that V2nd is correctby showing that a matrix element of V2nd with general, properly symmetrized, n particle states|Φ′〉 and |Φ〉 (i.e., 〈Φ′|V2nd|Φ〉) is the same as the matrix element of the same states with the1st quantization operator for n identical particles: i.e.,

V1st =1

2

∑

ij,i6=j

V (~ri − ~rj) ,

where the 1/2 prevents double counting. HINTS: You should recall the effect of a field creationoperator on a localized state of n− 1 particles: i.e.,

Ψs(~r )†|~r1s1, . . . , ~rn−1sn−1〉 =√n|~r1s1, . . . , rn−1sn−1, rnsn〉

(Ba-419) Also recall localized state unit operator for a properly symmetrized states of n particlesis

1n =∑

s1...sn

∫

d~r1 . . . ~rn |~r1s1, . . . , ~rnsn〉〈~r1s1, . . . , ~rnsn|

(Ba-421).

024 qfull 00500 3 5 0 tough thinking: fermion Coulomb exchangeExtra keywords: See Ba-436

6. The exchange energy per electron for a ground state electron gas is given by

E

N= −9πne2

p2f

∫ ∞

0

[sin(x) − x cos(x)]2

x5dx ,

where N is electron number, n is electron density, pf is the Fermi momentum, and x = pfr is adimensionless radius for an integration over all space (Ba-436).

a) First, let us analyze the integrand of the integral. Where are its zeros approximately forx ≥ 0? What is its small x behavior to 5th order in small x? What is its large x behavior?Sketch the integrand.

b) Now evaluate the integral approximately or exactly by analytic means. HINT: An exactanalytic integration must be possible, but probably one must use some special method. Forapproximate analytic integration, just do the best you can.

c) Now evaluate the integral numerically to high accuracy. HINT: Simpson’s rule with doubleprecision fortran works pretty well.


024 qfull 00600 2 5 0 moderate thinking: two fermions in a boxExtra keywords: (Ba-439:4)

7. Consider a box quantization system with volume V : the single-particle eigenstates recall aregiven by

φs(~r ) =ei~p·~r

V,

where periodic boundary conditions have been imposed, h− has been set to 1, and s gives thespin coordinate. We will consider two spin 1/2 fermions in the box in state

|1~p1s1, 1~p2s2

〉 = a†~p2s2a†~p1s1

|0〉 ,

where the a†’s are creation operators and |0〉 is the vacuum state (Ba-417). Recall theanticommutation relations for the fermion creation and annihilation operators:

a~ps, a†~p′s′ = δ~p~p′δss′ , a~ps, a~p′s′ = 0 , and a†~ps, a

†~p′s′ = 0 ,

where the a†’s are again creation operators and the a’s are annihilation operators (Ba-417).

a) Show that〈1~p1s1

, 1~p2s2|1~p1s1

, 1~p2s2〉 = 1 − δ~p1~p2

δs1s2.

What is the interpretation of this result?

b) You are now given the operator

a†~psa†~qs′a~q ′s′a~p ′s .

Write out the explicit expectation value of the operator for the |1~p1s1, 1~p2s2

〉 state. HINTS:Remember to make use of the annihilation property a|0〉 = 0. There are a lot of tediousKronecker deltas.

c) One form of the 2nd quantization two-body interaction operator is

v2nd =1

2

1

V 2

∑

pp′qq′

∑

ss′

∫

d~r d~r ′ v(~r − ~r ′)e−i(~p−~p ′)·~re−i(~q−~q ′)·~r ′

a†~psa†~qs′a~q ′s′a~p ′s ,

where V again is volume, v(~r−~r ′) is the two-body potential function, and a†~psa†~qs′a~q ′s′a~p ′s

is the operator from the part (b) question. Making use of the part (b) answer expressexpectation value 〈1~p1s1

, 1~p2s2|v2nd|1~p1s1

, 1~p2s2〉 as simply as possible. Don’t assume that

surface effects can be neglected. What is the use or significance of this matrix element?HINTS: The summations allow one to kill most of the Kroneckar deltas and get a prettysimple expression, but the explicit integrations are still there.

d) Say the two fermions are in different spin states and the two-body potential v(~r−~r ′) = C,where C is a constant. What is the expectation value from the part (c) answer in this case?

Chapt. 25 Klein-Gordon Equation


025 qmult 00100 1 4 4 easy deducto-memory: Klein-Gordan eqnExtra keywords: See the biography of Schrodinger and BJ-343

1. “Let’s play Jeopardy! For $100, the answer is: A relativistic quantum mechanical particleequation originally invented by Schrodinger (but never published by him) whose validinterpretation was provided by Pauli and Weisskopf.”

a) What is the 2nd Schrodinger equation, Alex?b) What is the relativistic Schodinger equation, Alex?c) What is the Pauli-Weisskopf equation, Alex?d) What is the Klein-Gordon equation, Alex?e) What is the small Scotsman equation, Alex?

025 qmult 00200 2 1 2 moderate memory: KG eqn developmentExtra keywords: See Ba-501

2. To develop the free-particle Klein-Gordon equation one assumes that the operators Eop =ih−∂/∂t and pop = (h−/i)∇ apply in relativistic quantum mechanics and then applies thecorrespondence principle to the non-quantum-mechanical special relativity result:

a) E =√

(pc)2 + (mc2)2.b) E2 = (pc)2 + (mc2)2.

c) γ = 1/√

1 − β2.d) E = mc2.

e) ℓ = ℓproper

√

1 − β2.

025 qmult 00300 1 1 2 easy memory: antiparticlesExtra keywords: See Ba-506

3. The free-particle Klein-Gordon equation leads to two energy eigen-solutions for each momentumeigenvalue. The two energies are equal in absolute value and opposite in sign in oneinterpretation. The negative energy solution can, in fact, be interpreted as a positive energysolution for:

a) a helium atom.b) an antiparticle.c) a wavicle.d) an anti-wavicle.e) an anti-wastrel.

025 qmult 00400 1 1 4 easy memory: KG density and current densityExtra keywords: See Ba-503 and BJ-343

4. The density and current density that one obtains from the Klein-Gordon wave function areinterpreted, respectively, as:

a) anti-density and anti-current density.

159

160 Chapt. 25 Klein-Gordon Equation

b) improbability density and current density.c) current density and density in a relativistic reversal.d) expectation value charge density and charge current density.e) probability density and probability current density.


025 qfull 00300 2 5 0 moderate thinking: Lorentz transf. of KG eqnExtra keywords: See Ba-502

1. One form of the free-particle Klein-Gordon (KG) equation is

KopΨ(~r, t) =

[

1

c2∂2

∂t2−∇2 +

(

mc

h−

)2]

Ψ(~r, t) = 0 ,

where Kop is here just an abbreviation for the Klein-Gordon equation operator and m is restmass (Ba-501). This equation is supposed to be relativistically correct. This means that it mustbe the correct physics to apply to the particle in any inertial frame. But how does the solutionΨ(~r, t) change on transformation from one inertial frame to another?

To find out, consider the special case of a transformation between frames S and S′, wherethe S′ is moving at velocity β (in units of c) along the mutual x-axis of the two frames. Theorigins of the two frames were coincident at τ = τ ′ = 0 (where τ = ct: i.e., time in units ofdistance). The special Lorentz transformations in this case are

x′ = γ(x− βτ) , τ ′ = γ(τ − βx) , y′ = y , and z′ = z ,

where the Lorentz factor

γ =1

√

1 − β2.

Now transform KopΨ(~r, t) = 0 to the primed system (i.e., write it terms of the ~r ′ and t′

variables). Does the transformed equation K ′opΨ[~r = f(~r ′, t′), t = f(~r ′, t′)] = 0 have the same

form as the untransformed equation: i.e., does it look like

[

1

c2∂2

∂t′2−∇′2 +

(

mc

h−

)2]

Ψ(~r ′, t′)′ = 0

when Ψ[~r = f(~r ′, t′), t = f(~r ′, t′)] is identified as Ψ(~r ′, t′)′? What kind of object is Ψ(~r, t): i.e.,scalar, vector, tensor? HINT: Recall the chain rule.



030 qfull 00100 2 5 0 moderate thinking: eph Bose-Einstein condensateExtra keywords: Greiner et al. 2002, 415, 39 with Stoof review on p. 25

Chapt. 25 Klein-Gordon Equation 161

1. Go to the library 2nd floor reading room and find the 20002jan03 issue of Nature (it may havebeen placed under the display shelf) and read the commentary article by Stoof on page 25 abouta quantum phase transition from a Bose-Einstein condensate to a Mott insulator. What do youthink Stoof really means when he says in the superfluid state “atoms still move freely from onevalley to the next”? NOTE: The instructor disavows any ability to completely elucidate thiscommentary or the research article by Greiner et al. it comments on.

030 qfull 00200 2 3 0 moderate thinking: eph quantum gravity well

Extra keywords: reference: Nesvizhevsky et al. 2002, Nature, 413, 297

2. Is the gravity subject to quantum mechanical laws or is it somehow totally decoupled? Everyonereally assumes that gravity is subject to quantum mechanical laws, but the assumption is notwell verified experimentally: in fact it may never have been verified at all until now—not thatI would know. The lack of experimental verification is because gravity is so infernally weakcompared to other forces in microscopic experiments that it is usually completely negligible.Recently Nesviszhevsky et al. (2002, Nature, 413, 297) have reported from an experiment thata gravity well (at least part of the constraining potential is gravitational) does have quantizedenergy states. This appears to be the first time that such an experimental result has beenachieved. It’s a wonderful result. Of course, if they hadn’t found quantization, it would havebeen a shock and most people would have concluded that the experiment was wrong somehow.Experiment may be the ultimate judge of theory, but experiment can certainly tell fibs forawhile.

Go read Nesviszhevsky et al. in the 2nd floor reading room of the library: the relevantissue may be under the shelf. If a neutron in the theoretically predicted gravity well made atransition from the 1st excited state to the ground state and emitted a photon, what would bethe wavelength of the photon? Could such a photon be measured? What classically does sucha transition correspond to?

030 qfull 00300 1 5 0 easy thinking: eph quantum computing

Extra keywords: Reference Seife, C. 2001, Science, 293, 2026

3. Read the article on quantum computing by Seife (2001, Science, 293, 2026) and make an estimateof how long it will be before for there is a quantum computer that solves a computationalproblem not solvable by a classical computer: I’m excepting, of course, any problems concerningquantum computer operation itself. Give your reasoning. All answers are right—and wrong—orin a superposition of those two states. My answer is 1 year. HINT: You can probably find theissue in the library, but there’s one in the physics lounge near the Britney issue. Primers onquantum computing can be found by going to

http://www.physics.unlv.edu/~jeffery/images/science

and clicking down through quantum mechanics and quantum computing.

030 qfull 00400 2 5 0 moderate thinking: eph C-70 diffraction

Extra keywords: Reference Nairz et al. 2001, quant-ph/0105061

4. On the web go to the Los Alamos eprint archive:

http://xxx.lanl.gov/ .

There click on search and then on search for articles by Zeilinger under the quant-ph topic.Locate Nairz, Arndt, & Zeilinger 2001, quant-ph/0105061 and download it. This is articlereports the particle diffraction for C70 (a fullerene). There should be great pictures on the webof fullerenes, but the best I could find were at

http://www.sussex.ac.uk/Users/kroto/fullgallery.html

and

http://cnst.rice.edu/pics.html

162 Chapt. 25 Klein-Gordon Equation

and these don’t have descriptions. Fullerenes are the largest particles ever shown to diffract:their size scale must be order-of a nanometer: 10 times ordinary atomic size. The article callsitself a verification of the Heisenberg uncertainty principle. In a general sense this is absolutelytrue since they verify the wave nature of particle propagation. But it isn’t a direct test of theformal uncertainty relation

σxσpx≥ h−

2,

where σx and σpxare standard deviations of x-direction position and momentum, respectively,

for the wave function (e.g., Gr-18, Gr-108–110). Explain why it isn’t a direct test. HINTS:You should all have studied physical optics at some point. Essentially what formula are theytesting?

030 qfull 00500 1 5 0 easy thinking: bulk and branesExtra keywords: Reference Arkani-Hamed, N., et al. 2002, Physics Today, February, 35

5. Go the library basement and read the article by Arkani-Hamed et al. (Physics Today, February,p. 35). Perhaps a 2nd reading would help or a course in particle physics. Anyway what is thebulk (not Hulk, bulk) and the brane (not Brain, brane)?

030 qfull 00600 1 5 0 easy thinking: sympathetic coolingExtra keywords: O’Hara & Thomas, 2001, Science, March 30, 291, 2556

6. Go to the library or the physics lounge and read O’Hara & Thomas ( 2001, Science, March 30,291, p. 2556) on degenerate gases of bosons and fermions. What is sympathetic cooling?

Appendix 1 Mathematical Problems



031 qfull 00100 2 3 0 moderate math: Gauss summations1. Gauss at the age of two proved various useful summation formulae. Now we can do this too

maybe.

a) Prove

S0(n) =

n∑

ℓ=1

1 = n .

HINT: This is really very easy.

b) Prove

S1(n) =

n∑

ℓ=1

ℓ =n(n+ 1)

2.

HINT: The trick is to add to every term in the sum its “complement” and then sum those2-sums and divide by 2 to account for double counting.

c) Prove

S2(n) =

n∑

ℓ=1

ℓ2 =n(n+ 1)(2n+ 1)

6.

HINT: A proof by induction works, but for that proof you need to know the result firstand that’s the weak way. The stronger way is to reduce the problem to an already solvedproblem. Consider the general summation formula

Sk(n) =

n∑

ℓ=1

ℓk .

For each ℓ, you can construct a column of ℓk−1’s that is ℓ in height. Can you add up thevalues in the table that is made up of these columns in some way to get Sk(n).

d) Prove

S3(n) =n∑

ℓ=1

ℓ3 =n2(n+ 1)2

4.

HINT: This formula can be proven using the “complement” trick and the formulae ofparts (b) and (c). It can also more tediously be solved by the procedure hinted at inpart (c). Or, of course, induction will work.

163

164 Appendix 1 Mathematical Problems

031 qfull 00200 1 3 0 easy math: uniqueness of power series2. Power series are unique.

a) Prove that coefficients ak of the power series

P (x) =

∞∑

k=0

akxk

are unique choices given that the series is convergent of course. HINT: The mth derivativeof P (x) evaluated at x = 0 can have only one value.

b) Prove that coefficients akℓ of the double power series

P (x, y) =

∞,∞∑

k=0,ℓ=0

akℓxkyℓ

are unique choices given that the series is convergent of course. HINT: Mutatis mutandis.

031 qfull 00300 2 5 0 moderate thinking: Leibniz’s formula (Ar-558) proof3. Prove Leibniz’s formula (Ar-558)

dn(fg)

dxn=

n∑

k=0

(

n

k

)

dkf

dxk

dn−kg

dxn−k

by induction.

031 qfull 00400 2 5 0 moderate thinking: integrals of type xe**-24. In evaluating anything that depends on a Gaussian distribution (e.g., the Maxwell-Boltzmann

distribution of classical statistical mechanics), one frequently has to evaluate integrals of thetype

In =

∫ ∞

0

xne−λx2

dx ,

where n is an odd positive integer.

a) Solve for I1.

b) Obtaining the general formula for In is now trivial with a magic trick. Act on I1 with theoperator

(

− d

dλ

)(n−1)/2

.

c) From the general formula evaluate I1 I3, I5, and I7.

031 qfull 01000 2 5 0 moderate thinking:5. In understanding determinants some permutation results must be proven. The proofs are

expected to be cogent and memorable rather than mathematical rigorous.

a) Given n objects, prove that there are n! permutations for ordering them in a line.

b) If you interchange any two particles in a given permutation, you get another permutation.Let’s call that action an exchange. If you exchange nearest nearest neighbors, let’s call thata nearest neighbor or NN exchange. Prove that any exchange requires an odd number ofNN exchanges.

Appendix 1 Mathematical Problems 165

c) Permutations have definite parity. This means that going from one definite permutation toanother definite permutation by any possible series of NN exchanges (i.e., by any possiblepath) will always involve either an even number of NN exchanges or an odd number: i.e., ifone path is even/odd, then any other path is even/odd. Given that definite parity is trueprove that any path of NN exchanges from a permutation that brings you back to thatpermutation (i.e., a closed path) has an even number of NN exchanges.

d) Now we have to prove definite parity exists. Say there is a fiducial permutation which bydefinition we say has even parity. If definite parity exists, then every other permutation isdefinitely even or odd relative to the fiducial permutation. If n = 1, does definite parityhold in this trivial case? For n ≥ 2, prove that definite parity holds. HINTS: It suffices toprove that going from the fiducial permutation to any other permutation always involvesa definite even or odd path since the fiducial permutation is arbitrary. Proof by inductionmight be the best route. I can’t see how brief word arguments can be avoided.

e) Now prove for n ≥ 2 that there are an equal number of even and odd permutations. HINT:Consider starting with an even permutation and systematically by an NN exchange pathgoing through all possible permutations. Then start with an odd permutation and followthe same NN exchange path.

Appendix 2 Quantum Mechanics Equation Sheet

Note: This equation sheet is intended for students writing tests or reviewing material. Thereforeit neither intended to be complete nor completely explicit. There are fewer symbols than variables,and so some symbols must be used for different things.

1 Constants not to High Accuracy

Constant Name Symbol Derived from CODATA 1998

Bohr radius aBohr =λCompton

2πα= 0.529 A

Boltzmann’s constant k = 0.8617× 10−6 eV K−1

= 1.381× 10−16 erg K−1

Compton wavelength λCompton =h

mec= 0.0246 A

Electron rest energy mec2 = 5.11 × 105 eV

Elementary charge squared e2 = 14.40 eV A

Fine Structure constant α =e2

h−c= 1/137.036

Kinetic energy coefficienth−2

2me= 3.81 eV A2

h−2

me= 7.62 eV A2

Planck’s constant h = 4.15 × 10−15 eVPlanck’s h-bar h− = 6.58 × 10−16 eV

hc = 12398.42 eVAh−c = 1973.27 eVA

Rydberg Energy ERyd =1

2mec

2α2 = 13.606 eV

2 Some Useful Formulae

Leibniz’s formuladn(fg)

dxn=

n∑

k=0

(

n

k

)

dkf

dxk

dn−kg

dxn−k

Normalized Gaussian P =1

σ√

2πexp

[

− (x− 〈x〉)22σ2

]

3 Schrodinger’s Equation

HΨ(x, t) =

[

p2

2m+ V (x)

]

Ψ(x, t) = ih−∂Ψ(x, t)

∂t

Hψ(x) =

[

p2

2m+ V (x)

]

ψ(x) = Eψ(x)

HΨ(~r , t) =

[

p2

2m+ V (~r )

]

Ψ(~r , t) = ih−∂Ψ(~r , t)

∂tH |Ψ〉 = ih− ∂

∂t|Ψ〉

166

Appendix 2 Quantum Mechanics Equation Sheet 167

Hψ(~r ) =

[

p2

2m+ V (~r )

]

ψ(~r ) = Eψ(~r ) H |ψ〉 = E|ψ〉

4 Some Operators

p =h−i

∂

∂xp2 = −h−2 ∂2

∂x2

H =p2

2m+ V (x) = − h−2

2m

∂2

∂x2+ V (x)

p =h−i∇ p2 = −h−2∇2

H =p2

2m+ V (~r ) = − h−2

2m∇2 + V (~r )

∇ = x∂

∂x+ y

∂

∂y+ z

∂

∂z= r

∂

∂r+ θ

1

r

∂

∂θ+ θ

1

r sin θ

∂

∂θ

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2=

1

r2∂

∂r

(

r2∂

∂r

)

+1

r2 sin θ

∂

∂θ

(

sin θ∂

∂θ

)

+1

r2 sin2 θ

∂2

∂φ2

5 Kronecker Delta and Levi-Civita Symbol

δij =

1, i = j;0, otherwise

εijk =

1, ijk cyclic;−1, ijk anticyclic;0, if two indices the same.

εijkεiℓm = δjℓδkm − δjmδkℓ (Einstein summation on i)

6 Time Evolution Formulae

Generald〈A〉dt

=

⟨

∂A

∂t

⟩

+1

h−〈i[H(t), A]〉

Ehrenfest’s Theoremd〈~r 〉dt

=1

m〈~p 〉 and

d〈~p 〉dt

= −〈∇V (~r )〉

|Ψ(t)〉 =∑

j

cj(0)e−iEjt/h−|φj〉

7 Simple Harmonic Oscillator (SHO) Formulae

V (x) =1

2mω2x2

(

− h−2

2m

∂2

∂x2+

1

2mω2x2

)

ψ = Eψ

168 Appendix 2 Quantum Mechanics Equation Sheet

β =

√

mω

h−ψn(x) =

β1/2

π1/4

1√2nn!

Hn(βx)e−β2x2/2 En =

(

n+1

2

)

h−ω

H0(βx) = H0(ξ) = 1 H1(βx) = H1(ξ) = 2ξ

H2(βx) = H2(ξ) = 4ξ2 − 2 H3(βx) = H3(ξ) = 8ξ3 − 12ξ

8 Position, Momentum, and Wavenumber Representations

p = h−k Ekinetic = ET =h−2k2

2m

|Ψ(p, t)|2 dp = |Ψ(k, t)|2 dk Ψ(p, t) =Ψ(k, t)√

h−

xop = x pop =h−i

∂

∂xQ

(

x,h−i

∂

∂x, t

)

position representation

xop = − h−i

∂

∂ppop = p Q

(

− h−i

∂

∂p, p, t

)

momentum representation

δ(x) =

∫ ∞

−∞

eipx/h−

2πh−dp δ(x) =

∫ ∞

−∞

eikx

2πdk

Ψ(x, t) =

∫ ∞

−∞

Ψ(p, t)eipx/h−

(2πh−)1/2dp Ψ(x, t) =

∫ ∞

−∞

Ψ(k, t)eikx

(2π)1/2dk

Ψ(p, t) =

∫ ∞

−∞

Ψ(x, t)e−ipx/h−

(2πh−)1/2dx Ψ(k, t) =

∫ ∞

−∞

Ψ(x, t)e−ikx

(2π)1/2dx

Ψ(~r , t) =

∫

all space

Ψ(~p , t)ei~p·~r/h−

(2πh−)3/2d3p Ψ(~r , t) =

∫

all space

Ψ(~k , t)ei~k·~r

(2π)3/2d3k

Ψ(~p , t) =

∫

all space

Ψ(~r , t)e−i~p·~r/h−

(2πh−)3/2d3r Ψ(~k , t) =

∫

all space

Ψ(~r , t)e−i~k·~r

(2π)3/2d3r

9 Commutator Formulae

[A,BC] = [A,B]C +B[A,C]

∑

i

aiAi,∑

j

bjBj

=∑

i,j

aibj [Ai, bj]

if [B, [A,B]] = 0 then [A,F (B)] = [A,B]F ′(B)

[x, p] = ih− [x, f(p)] = ih−f ′(p) [p, g(x)] = −ih−g′(x)


[a, a†] = 1 [N, a] = −a [N, a†] = a†

10 Uncertainty Relations and Inequalities

σxσp = ∆x∆p ≥ h−2

σQσQ = ∆Q∆R ≥ 1

2|〈i[Q,R]〉|

σH∆tscale time = ∆E∆tscale time ≥h−2

11 Probability Amplitudes and Probabilities

Ψ(x, t) = 〈x|Ψ(t)〉 P (dx) = |Ψ(x, t)|2 dx ci(t) = 〈φi|Ψ(t)〉 P (i) = |ci(t)|2

12 Spherical Harmonics

Y0,0 =1√4π

Y1,0 =

(

3

4π

)1/2

cos(θ) Y1,±1 = ∓(

3

8π

)1/2

sin(θ)e±iφ

L2Yℓm = ℓ(ℓ+ 1)h−2Yℓm LzYℓm = mh−Yℓm |m| ≤ ℓ m = −ℓ,−ℓ+ 1, . . . , ℓ− 1, ℓ

0 1 2 3 4 5 6 . . .s p d f g h i . . .

13 Hydrogenic Atom

ψnℓm = Rnℓ(r)Yℓm(θ, φ) ℓ ≤ n− 1 ℓ = 0, 1, 2, . . . , n− 1

az =a

Z

(

me

mreduced

)

a0 =h−

mecα=

λC

2παα =

e2

h−c

R10 = 2a−3/2Z e−r/aZ R20 =

1√2a−3/2Z

(

1 − 1

2

r

aZ

)

e−r/(2aZ)

R21 =1√24a−3/2Z

r

aZe−r/(2aZ)

Rnℓ = −

(

2

naZ

)3(n− ℓ− 1)!

2n[(n+ ℓ)!]3

1/2

e−ρ/2ρℓL2ℓ+1n+ℓ (ρ) ρ =

2r

nrZ

Lq(x) = ex

(

d

dx

)q(

e−xxq)

Rodrigues’s formula for the Laguerre polynomials

Ljq(x) =

(

d

dx

)j

Lq(x) Associated Laguerre polynomials


〈r〉nℓm =aZ

2

[

3n2 − ℓ(ℓ+ 1)]

Nodes = (n− 1) − ℓ not counting zero or infinity

En = −1

2mec

2α2Z2

n2

mreduced

me= −ERyd

Z2

n2

mreduced

me= −13.606

Z2

n2

mreduced

meeV

14 General Angular Momentum Formulae

[Ji, Jj ] = ih−εijkJk (Einstein summation on k) [J2, ~J ] = 0

J2|jm〉 = j(j + 1)h−2|jm〉 Jz |jm〉 = mh−|jm〉

J± = Jx ± iJy J±|jm〉 = h−√

j(j + 1) −m(m± 1)|jm± 1〉

Jx

y =

1212i

(J+ ± J−) J†±J± = J∓J± = J2 − Jz(Jz ± h−)

[Jfi, Jgj ] = δfgih−εijkJk~J = ~J1 + ~J2 J2 = J2

1 + J22 + J1+J2− + J1−J2+ + 2J1zJ2z

J± = J1± + J2± |j1j2jm〉 =∑

m1m2,m=m1+m2

|j1j2m1m2〉〈j1j2m1m2|j1j2jm〉j1j2jm〉

|j1 − j2| ≤ j ≤ j1 + j2

j1+j2∑

|j1−j2|

(2j + 1) = (2j1 + 1)(2j2 + 1)

15 Spin 1/2 Formulae

Sx =h−2

(

0 11 0

)

Sy =h−2

(

0 −ii 0

)

Sz =h−2

(

1 00 −1

)

|±〉x =1√2

(|+〉 ± |−〉) |±〉y =1√2

(|+〉 ± i|−〉) |±〉z = |±〉

| + +〉 = |1,+〉|2,+〉 | + −〉 =1√2

(|1,+〉|2,−〉± |1,−〉|2,+〉) | − −〉 = |1,−〉|2,−〉

σx =

(

0 11 0

)

σy =

(

0 −ii 0

)

σz =

(

1 00 −1

)


σiσj = δij + iεijkσk [σi, σj ] = 2iεijkσk σi, σj = 2δij

( ~A · ~σ)( ~B · ~σ) = ~A · ~B + i( ~A× ~B) · ~σ

d(~S · n)

dα= − i

h−[~S · α, ~S · n] ~S · n = e−i~S·~α~S · n0e

i~S·~α |n±〉 = e−i~S·~α|z±〉

eixA = 1 cos(x) + iA sin(x) if A2 = 1 e−i~σ·~α/2 = 1 cos(x) − i~σ · α sin(x)

σif(σj) = f(σj)σiδij + f(−σj)σi(1 − δij)

µBohr =eh−2m

= 0.927400915(23)× 10−24 J/T = 5.7883817555(79)× 10−5 eV/T

g = 2(

1 +α

2π+ . . .

)

= 2.0023193043622(15)

~µorbital = −µBohr

~L

h−~µspin = −gµBohr

~S

h−~µtotal = ~µorbital + ~µspin = −µBohr

(~L + g~S)

h−

Hµ = −~µ · ~B Hµ = µBohrBz(Lz + gSz)

h−

16 Time-Independent Approximation Methods

H = H(0) + λH(1) |ψ〉 = N(λ)

∞∑

k=0

λk|ψ(k)n 〉

H(1)|ψ(m−1)n 〉(1 − δm,0) +H(0)|ψ(m)

n 〉 =

m∑

ℓ=0

E(m−ℓ)|ψ(ℓ)n 〉 |ψ(ℓ>0)

n 〉 =

∞∑

m=0, m 6=n

anm|ψ(0)n 〉

|ψ1stn 〉 = |ψ(0)

n 〉 + λ∑

all k, k 6=n

⟨

ψ(0)k |H(1)|ψ(0)

n

⟩

E(0)n − E

(0)k

|ψ(0)k 〉

E1stn = E(0)

n + λ⟨

ψ(0)n |H(1)|ψ(0)

n

⟩

E2ndn = E(0)

n + λ⟨

ψ(0)n |H(1)|ψ(0)

n

⟩

+ λ2∑

all k, k 6=n

∣

∣

∣

⟨

ψ(0)k |H(1)|ψ(0)

n

⟩∣

∣

∣

2

E(0)n − E

(0)k

E(φ) =〈φ|H |φ〉〈φ|φ〉 δE(φ) = 0


Hkj = 〈φk|H |φj〉 H~c = E~c

17 Time-Dependent Perturbation Theory

π =

∫ ∞

−∞

sin2(x)

x2dx

Γ0→n =2π

h−|〈n|Hperturbation|0〉|2δ(En − E0)

18 Interaction of Radiation and Matter

~Eop = −1

c

∂ ~Aop

∂t~Bop = ∇× ~Aop

19 Box Quantization

kL = 2πn, n = 0,±1,±2, . . . k =2πn

L∆kcell =

2π

L∆k3

cell =(2π)3

V

dNstates = gk2 dk dΩ

(2π)3/V

20 Identical Particles

|a, b〉 =1√2

(|1, a; 2, b〉 ± |1, b; 2, a〉)

ψ(~r1, ~r 2) =1√2

(ψa(~r 1)ψb(~r 2) ± ψb(~r 1)ψa(~r 2))

21 Second Quantization

[ai, a†j ] = δij [ai, aj ] = 0 [a†i , a

†j ] = 0 |N1, . . . , Nn〉 =

(a†n)Nn

√Nn!

. . .(a†1)

N1

√N1!

|0〉

ai, a†j = δij ai, aj = 0 a†i , a

†j = 0 |N1, . . . , Nn〉 = (a†n)Nn . . . (a†1)

N1 |0〉

Ψs(~r )† =∑

~p

e−i~p·~r

√V

a†~ps Ψs(~r ) =∑

~p

ei~p·~r

√Va~ps

[Ψs(~r ),Ψs′(~r ′)]∓ = 0 [Ψs(~r )†,Ψs′(~r ′)†]∓ = 0 [Ψs(~r ),Ψs′(~r ′)†]∓ = δ(~r − ~r ′)δss′


|~r1s1, . . . , ~rnsn〉 =1√n!

Ψsn(~r n)† . . .Ψsn

(~r n)†|0〉

Ψs(~r )†|~r1s1, . . . , ~rnsn〉√n+ 1|~r1s1, . . . , ~rnsn, ~rs〉

|Φ〉 =

∫

d~r1 . . . d~rn Φ(~r1, . . . , ~rn)|~r1s1, . . . , ~rnsn〉

1n =∑

s1...sn

∫

d~r1 . . . d~rn |~r1s1, . . . , ~rnsn〉〈~r1s1, . . . , ~rnsn| 1 = |0〉〈0| +∞∑

n=1

1n

N =∑

~ps

a†~psa~ps T =∑

~ps

p2

2ma†~psa~ps

ρs(~r ) = Ψs(~r )†Ψs(~r ) N =∑

s

∫

d~r ρs(~r ) T =1

2m

∑

s

∫

d~r∇Ψs(~r )† · ∇Ψs(~r )

~js(~r ) =1

2im

[

Ψs(~r )†∇Ψs(~r ) − Ψs(~r )∇Ψs(~r )†]

Gs(~r − ~r ′) =3n

2

sin(x) − x cos(x)

x3gss′(~r − ~r ′) = 1 − δss′

Gs(~r − ~r ′)2

(n/2)2

v2nd =1

2

∑

ss′

∫

d~rd~r ′ v(~r − ~r ′)Ψs(~r )†Ψs′(~r ′)†Ψs′(~r ′)Ψs(~r )

v2nd =1

2V

∑

pp′qq′

∑

ss′

v~p−~p ′δ~p+~q,~p′+~q′a†~psa†~qs′a~q ′s′a~p ′s v~p−~p ′ =

∫

d~r e−i(~p−~p ′)·~rv(~r )

22 Klein-Gordon Equation

E =√

p2c2 +m2c41

c2

(

ih− ∂

∂t

)2

Ψ(~r, t) =

[

(

h−i∇)2

+m2c2

]

Ψ(~r, t)

[

1

c2∂2

∂t2−∇2 +

(

mc

h−

)2]

Ψ(~r, t) = 0

ρ =ih−

2mc2

(

Ψ∗ ∂Ψ

∂t− Ψ

∂Ψ∗

∂t

)

~j =h−

2im(Ψ∗∇Ψ − Ψ∇Ψ∗)

1

c2

(

ih− ∂

∂t− eΦ

)2

Ψ(~r, t) =

[

(

h−i∇− e

c~A

)2

+m2c2

]

Ψ(~r, t)

Ψ+(~p,E) = ei(~p·~r−Et)/h− Ψ−(~p,E) = e−i(~p·~r−Et)/h−

Appendix 3 Multiple-Choice Problem Answer Tables

Note: For those who find scantrons frequently inaccurate and prefer to have their own table andmarking template, the following are provided. I got the template trick from Neil Huffacker atUniversity of Oklahoma. One just punches out the right answer places on an answer table andoverlays it on student answer tables and quickly identifies and marks the wrong answers

Answer Table for the Multiple-Choice Questions

a b c d e a b c d e

1. O O O O O 6. O O O O O




5. O O O O O 10. O O O O O

174

Appendix 3 Multiple-Choice Problem Answer Tables 175


a b c d e a b c d e

1. O O O O O 11. O O O O O

2. O O O O O 12. O O O O O

3. O O O O O 13. O O O O O

4. O O O O O 14. O O O O O

5. O O O O O 15. O O O O O

6. O O O O O 16. O O O O O

7. O O O O O 17. O O O O O

8. O O O O O 18. O O O O O

9. O O O O O 19. O O O O O

10. O O O O O 20. O O O O O

176 Appendix 3 Multiple-Choice Problem Answer Tables


a b c d e a b c d e

1. O O O O O 16. O O O O O

2. O O O O O 17. O O O O O

3. O O O O O 18. O O O O O

4. O O O O O 19. O O O O O

5. O O O O O 20. O O O O O

6. O O O O O 21. O O O O O

7. O O O O O 22. O O O O O

8. O O O O O 23. O O O O O

9. O O O O O 24. O O O O O

10. O O O O O 25. O O O O O

11. O O O O O 26. O O O O O

12. O O O O O 27. O O O O O

13. O O O O O 28. O O O O O

14. O O O O O 29. O O O O O

15. O O O O O 30. O O O O O

Appendix 3 Multiple-Choice Problem Answer Tables 177

NAME:


a b c d e a b c d e

1. O O O O O 21. O O O O O

2. O O O O O 22. O O O O O

3. O O O O O 23. O O O O O

4. O O O O O 24. O O O O O

5. O O O O O 25. O O O O O

6. O O O O O 26. O O O O O

7. O O O O O 27. O O O O O

8. O O O O O 28. O O O O O

9. O O O O O 29. O O O O O

10. O O O O O 30. O O O O O

11. O O O O O 31. O O O O O

12. O O O O O 32. O O O O O

13. O O O O O 33. O O O O O

14. O O O O O 34. O O O O O

15. O O O O O 35. O O O O O

16. O O O O O 36. O O O O O

17. O O O O O 37. O O O O O

18. O O O O O 38. O O O O O

19. O O O O O 39. O O O O O

20. O O O O O 40. O O O O O

178 Appendix 3 Multiple-Choice Problem Answer Tables

Answer Table Name:a b c d e a b c d e

1. O O O O O 31. O O O O O

2. O O O O O 32. O O O O O

3. O O O O O 33. O O O O O

4. O O O O O 34. O O O O O

5. O O O O O 35. O O O O O

6. O O O O O 36. O O O O O

7. O O O O O 37. O O O O O

8. O O O O O 38. O O O O O

9. O O O O O 39. O O O O O

10. O O O O O 40. O O O O O

11. O O O O O 41. O O O O O

12. O O O O O 42. O O O O O

13. O O O O O 43. O O O O O

14. O O O O O 44. O O O O O

15. O O O O O 45. O O O O O

16. O O O O O 46. O O O O O

17. O O O O O 47. O O O O O

18. O O O O O 48. O O O O O

19. O O O O O 49. O O O O O

20. O O O O O 50. O O O O O

21. O O O O O 51. O O O O O

22. O O O O O 52. O O O O O

23. O O O O O 53. O O O O O

24. O O O O O 54. O O O O O

25. O O O O O 55. O O O O O

26. O O O O O 56. O O O O O

27. O O O O O 57. O O O O O

28. O O O O O 58. O O O O O

29. O O O O O 59. O O O O O

30. O O O O O 60. O O O O O

Documents

Quantum Mechanics Problems David J. Jeffery